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Algebra and Trigonometry
Algebra
and
Trigonometry
ALVIN K. BETTINGER
Head of Department and
Professor of Mathematics
The Crcif/hton University
and
JOHN A. ENGLUND
Formerly Assistant Professor
of Mathematics
The Cr eight on University
INTERNATIONAL TEXTBOOK COMPANY
Scranton, Pennsylvania
INTERNATIONAL TEXTBOOKS IN MATHEMATICS
L R. W//COX
Professor of Mathematics
Illinois Institute of Technology
CONSULTING EDITOR
Second Printing, ] armory > 1963
Copyright I960, by International Textbook Company. All rights reserved.
Printed in the United States of America by The Haddon Craftsmen, Inc. t
at Scranton, Pennsylvania. Library of Congress Card Number: 60-9987.
Preface
The authors believe that in this book the basic material of college
algebra and trigonometry has been presented with suflicient rigor
to provide a firm and coherent groundwork for subsequent courses
in mathematics. The material is presented in such a way that it
can be grasped by the student without undue assistance.
The first chapter consists of a number of introductory topics
which are intended to serve as a review of elementary algebra.
Actually, something more than a mere review is available in this
chapter. Not only are the review topics considered from a more
mature point of view than is usual, but the treatment is inter-
woven with concepts that are basic for an understanding of more
advanced mathematical topics. We begin with the algebra of the
real-number system. Axioms pertaining to fundamental operations
are given, and the various rules for the elementary operations of
algebra are derived and logically connected with the bcisic assump-
tions. We are led naturally to an ordering of the real-number
system and to the foundation for a later chapter on inequalities
that is easier to understand and more useful than the treatment
one customarily finds in textbooks.
The second chapter introduces the student to the function con-
cept, which serves as a basis for much of the remaining work of
the book. Certain aspects of the discussion become somewhat
abstract, but the student is reminded that a proper understanding
of the true nature of a function is important for virtually all later
courses in mathematics.
In line with modern demands, the trigonometric functions are
initially introduced in the third chapter as functions of real num-
bers. Following this presentation, the transition to functions of
angles is relatively simple.
The rest of the volume contains all the usual topics from college
algebra and trigonometry. In certain instances a particular devel-
opment may differ somewhat from that usually found. In such
cases, the authors believe, the departure is to the advantage of
the student.
vi Preface
We are indebted to our colleague, Professor Morris Dansky, for
his valuable suggestions while the manuscript was in preparation.
We wish particularly to express our deep appreciation to Pro-
fessor L. R. Wilcox for his thorough criticism of the manuscript
and his invaluable suggestions for improvement of the text. Finally,
a special word of thanks is due the International Textbook Com-
pany for its cooperation and patience.
A. K. BETTINGER
J. A. ENGLUND
Omaha, Nebraska
August, 1960
Contents
1. INTRODUCTORY TOPICS 1
1-1. The Real-Number System 1
1-2. Fundamental Assumptions 1
1-3. Operations With Zero 5
1-4. Reciprocals 5
1-5. The Real-Number Scale 6
1-6. Rules of Signs 7
1-7. Fundamental Operations on Fractions 9
1-8. Order Relations for Real Numbers 12
1-9. Absolute Value 14
1-10. Inequalities Involving Absolute Values 14
1-11. Positive Integral Exponents 16
1-12. Algebraic Expressions 17
1-13. Equations and Identities 18
1-14. Symbols of Grouping 18
1-15. Order of Fundamental Operations 20
1-16. Addition and Subtraction of Algebraic Expressions 21
1-17. Multiplication of Algebraic Expressions 22
1-18. Special Products 22
1-19. Division of Algebraic Expressions 23
1-20. Factoring 25
1-21. Important Type Forms for Factoring 27
1-22. Greatest Common Divisor 30
1-23. Least Common Multiple 32
1-24. Reduction of Fractions 33
1-25. Signs Associated With Fractions 34
1-26. Addition and Subtraction of Fractions 36
1-27. Multiplication and Division of Fractions 39
1-28. Complex Fractions 40
1-29. Linear Equations 42
1-30. Linear Equations in One Unknown 43
2. THE FUNCTION CONCEPT 49
2-1. Rectangular Coordinate Systems in a Plane 49
2-2. Distance Between Two Points 50
2-3. Functions 52
2-4. Functional Notation 55
2-5. Some Special Functions 57
2-6. Variation 57
2-7. Classification of Functions 61
vii
viii Confenfs
3. THE TRIGONOMETRIC FUNCTIONS G3
3-1. The Point Function P(t) 63
3-2. Definitions of the Trigonometric Functions (54
3-3. Identities 68
3-4. Tables of Trigonometric Functions 71
3-5. Positive and Negative Angles and Standard Position 75
3-6. Measurement of Angles 76
3-7. The Relation Between Radians and Degrees 77
3-8. Arc Length and Area of a Sector 78
3-9. Trigonometric Functions of Angles 81
3-10. Tables of Natural Trigonometric Functions of Angles 82
4. THE LAWS OF EXPONENTS 86
4-1. Positive Integral Exponents 86
4-2. Meaning of 88
4-3. Negative Exponents DO
4-4. Scientific Notation 1)2
4-5. Rational Exponents 1)2
4-6. The Factorial Symbol J)7
4-7. The Binomial Theorem 1)7
48. General Term in the Binomial Expansion 1)9
5. LOGARITHMS 101
5-1. Definition of a Logarithm 101
5-2. Laws of Logarithms 102
5-3. Systems of Logarithms 105
5-4. Common Logarithms 105
5-5. Rules for Characteristic and Mantissa 106
5-6. How to Write Logarithms 108
5-7. How to Use a Table of Mantissas 108
5-8. Logarithmic Computation 110
5-9. Change of Base 113
6. RIGHT TRIANGLES AND VECTORS 115
6-1. Rounding Off Numbers 115
6-2. Trigonometric Functions of Acute Angles 116
6-3. Procedures for Solving Right Triangles 117
6-4. Angles of Elevation and Depression 120
6-5. Bearing in Navigation arid Surveying 121
6-6. Projections 122
6-7. Scalar and Vector Quantities 125
6-8. Logarithms of Trigonometric Functions 131
6-9. Logarithmic Solution of Right Triangles 133
7. TRIGONOMETRIC FUNCTIONS OF SUMS AND DIFFERENCES 135
7-1. Derivation of the Addition Formulas 135
7-2. The Double-Angle Formulas 139
7-3. The Half-Angle Formulas 140
7-4. Products of Two Functions Expressed as Sums, and Sums
Expressed as Products 143
Contents ix
8. GRAPHS OF TRIGONOMETRIC FUNCTIONS; INVERSE FUNCTIONS
AND THEIR GRAPHS 146
8-1. Variation of the Trigonometric Functions 146
8-2. The Graph of the Sine Function 117
8-3. The Graphs of the Cosine and Tangent Functions 148
8-4. Periodicity, Amplitude, and Phase 14i)
8-5. Inverse Functions ir>r>
8-6. Inverses of the Trigonometric Functions 156
9. LINEAR EQUATIONS AND GRAPHS 163
!)-!. Solutions of Simultaneous Equations 1(5*5
9-2. Algebraic Solution of Linear Equations in Two Unknowns... 166
9-3. Linear Equations in Three Unknown* 167
9-4. Graphs of Linear Functions 169
9-5. Intercepts 170
9-6. Graphical Solution of Linear Equations in Two Unknowns . . . 171
10. 1 )ETERMINANTS 173
10-1 . I )eti'rminants of the Second Order 173
10-2. Determinants of the Third Order 175
10-3. Properties of Determinants 177
10-4. Solution of Three Simultaneous Linear Equations in Three
Unknowns 181
10-5. Systems of Three Linear Equations in Three Unknowns When
D = () 183
10-6. Homogeneous Equations 184
10-7. Sum and Product of Determinants 185
1 1 . COMPLEX NUMBERS t 189
11-1. The Complex Number System ^ 189
11-2. The Standard Notation for Complex Numbers 191
11-3. Operations on Complex Numbers in Standard Form 192
11-4. Graphical Representation 105
11-5. Trigonometric Representation 106
11-6. Multiplication and Division in Trigonometric Form 198
11-7. DC Moivrc's Theorem 199
11-8. Roots of Complex Numbers 200
12. EQUATIONS IN QUADRATIC FORM 204
12-1. Quadratic Equations in One Unknown 204
1 2-2. Solution of Quadratic Equations by Factoring 20 1
12-3. Completing the Square 206
12-4. Solution of Quadratic Equations by the Quadratic Formula . . . 209
12-5. Equations Involving Radicals 212
12-6. Equations in Quadratic Form 214
12-7. The Discriminant 215
12-8. Sum and Product of the Roots 217
12-9. Graphs of Quadratic Functions 218
12-10. Quadratic Equations in Two Unknowns 221
x Confenfs
12-11. Graphical Solutions of Systems of Equations Involving
Quadratics 224
12-12. Algebraic Solutions of Systems Involving Quadratics 227
12-13. Exponential and Logarithmic Equations 231
12-14. Graphs of Logarithmic and Exponential Functions 233
13. THEORY OF EQUATIONS 235
1 3-1. Introductory Remarks 235
13-2. Synthetic Division 235
13-3. The Remainder Theorem 240
13-4. The Fundamental Theorem of Algebra 241
13-5. Pairs of Complex Roots of an Equation 243
13-6. The Graph of a Polynomial for Large Values of a- 244
13-7. Roots Between a and b If /(a) and f(b) Have Opposite Signs 245
13-8. Rational Roots 245
14. INEQUALITIES 248
14-1. Introduction 248
14-2. Properties of Inequalities 248
14-3. Solution of Conditional Inequalities 249
14-4. Absolute Inequalities 254
15. PROGRESSIONS 256
15-1 . Sequences and Series 256
15-2. Arithmetic Progressions 260
15-3. The General Term of an Arithmetic Progression 260
15-4. Sum of the First 71 Terms of an Arithmetic Progression 261
15-5. Arithmetic Means 262
15-6. Harmonic Progressions 264
1 5-7. Geometric Progression 265
15-8. The General Term of a Geometric Progression 265
15-9. Sum of the First n Terms of a Geometric Progression 266
15-10. Geometric Means 267
15-11. Infinite Geometric Progression 268
15-12. Repeating Decimals 269
15-13. The Binomial Series 271
16. MATHEMATICAL INDUCTION 273
16-1. Method of Mathematical Induction 273
16-2. Proof of the Binomial Theorem for Positive Integral Exponents 275
17. PERMUTATIONS, COMBINATIONS, AND PROBABILITY 278
17-1. Fundamental Principle 278
17-2. Permutations 279
17-3. Permutations of n Things Not All Different 280
17-4. Combinations 281
17-5. Binomial Coefficients 282
17-6. Mathematical Probability 283
17-7. Most Probable Number and Mathematical Expectation 284
17-8. Statistical, or Empirical, Probability 284
Confenfs xi
17-9. Mutually Exclusive Events 285
17-10. Dependent and Independent Events 286
17-11. Repeated Trials 287
18. SOLUTION OF THE GENERAL TRIANGLE 289
18-1. Classes of Problems 289
18-2. The Law of Sines 289
18-3. Solution of Case I by the Law of Sines: Given One Side and
Two Angles 290
18-4. Solution of Case II by the Law of Sines: Given Two Sides and
the Angle Opposite One of Them 291
18-5. The Law of Cosines 296
18-6. Solution of Case III and Case IV by the Law of Cosines 297
18-7. The Law of Tangents 298
18-8. The Half-Angle Formulas 300
18-9. Area of a Trjangle 302
APPENDIX
A. Tables 307
"B. Answers to Odd-Numbered Problems 337
INDEX 353
1
Introductory Topics
1-1. THE REAL-NUMBER SYSTEM
The real-number system that we use in the early part of this
course is a development from the original counting numbers, or
positive integers, such as 1, 2, and 3. Almost simultaneously with
the invention of positive integers, practical problems of measure-
ment gave rise to positive fractions, such as 1/2, 5/6, and 16/7.
Much later, in comparatively modern times, the concepts of negative
numbers and of other types of numbers were gradually developed.
Negative numbers were invented when the problem of subtracting
one number from a smaller one presented itself. Thus, the number
system was soon enlarged to include the negative integers and
fractions. These positive and negative numbers, together with zero,
are called the rational numbers. Hence, a rational number is
defined to be any number that can be expressed as the quotient, or
ratio, of two integers. For example, 2/3, 5 (which may be con-
sidered as 5/1), and 7 are rational numbers.
The number system was then extended to include also numbers
which cannot be expressed as the quotient of two integers, namely,
the irrational numbers; examples are ^/2 ami rr. The two classes of
numbers, rational and irrational, comprise the real numbers. These
numbers are so called in contrast to the imaginary or complex
numbers considered in Chapter 11.
1-2. FUNDAMENTAL ASSUMPTIONS
We shall proceed to introduce the four fundamental operations
of addition, subtraction, multiplication, and division into the system
of real numbers. The reader has probably been performing these
operations in arithmetic and algebra without being conscious that
certain basic laws were being obeyed. We shall introduce the four
fundamental operations and state, without proof, the laws or
assumptions governing them.
i
2 Introductory Topics Sec. 1-2
Addition. It is assumed that there is a mode of combining any
two real numbers a and 6 so as to produce a definite real number
called their sum. This mode of combination is called addition. The
sum of a and b is denoted by a + b. In this sum a and b are called
terms.
Multiplication. It is assumed that there is a mode of combining
any two real numbers a and b to produce a definite real number
called their product. This mode of combination is called multiplica-
tion. The product of a and b is denoted by a b or by ab. The individ-
ual numbers a and b are called factors of the product.
Commutative Law for Addition. If a and b are any real numbers,
then
(1-1) a + b = b + a.
Thus 1 , the sum of two mumbers is the same regardless of the order
in which they are added. For example,
2 + 3=3 + 2.
Associative Law for Addition. If a, b, c are any real numbers,
then
(1-2) (a + 6) + c = a + (6 + c).
That is, we obtain the same result whether we add the sum of a and
6 to c, or we add a to the sum of b and c. Since the way in which
we associate or group these numbers is immaterial, we may write
this common value as a + 6 4- c without fear of ambiguity. For
example,
2 + 3 + 4 = (2 + 3) + 4 = 2 + (3 + 4).
Commutative Law for Multiplication. If a and b are any real num-
bers, then
(1-3) ab - ba.
That is, the product of two numbers is the same regardless of the
order in which they are multiplied. For example,
2-3=3-2.
Associative Law for Multiplication. , If a, 6, c are any real num-
bers, then
(1-4) (o6)c = a(6c).
1 Illustrations of the laws are given here only for the most familiar num-
bers, the positive integers. It is understood, however, that the laws apply to
all real numbers.
Sec. 12 Introductory Topics 3
That is, we obtain the same result whether we multiply the product
of a and b by c, or we multiply a by the product of b and c. Since
the way in which we associate or group these numbers is imma-
terial, we may write the result as abc without fear of ambiguity.
Thus
2-:5-4 = (2-:}) i = 2- (:;-4).
Distributive Law. If a, b, c are any real numbers, then 2
(1 f>) a(b + r) = ah + or.
This law, which is usually known as the distributive law for multi-
plication with respect to addition, effects a connection between
addition and multiplication. The distributive law forms the basis
for the factoring process in algebra, as will be seen.
A simple example of the distributive law is
2 -(3 + 4) =2-:5 + 2- 1.
This law can be extended to the case where the sum consists of
three or more terms, as in the following illustration:
For positive integers, multiplication may also be interpreted as
repeated addition. Thus, by the distributive law,
S 4 = (1 + 1 + 1) - \ -- (1 -4) + (1 - 4) + (1 4) = 4 + 4 + 4,
;>,.4 = .">(! + 1 + I + 1) = (3- 1) + Cl- 1) + (o- 1) + C>- 1)
= :$ + :! + ;; + :j.
Zero. It is assumed that there is a special number called zero
and denoted by 0, such that, for every real number ,
(1 (i) a + = a.
For example,
i] + o=;j, o + i = j, o + o = o.
It can be easily shown that only one number with the property
of can exist. For let 0' be another such number. Then, since
a + = a and b + 0' = b for any numbers r/, b, it follows, by taking
a = 0' and b = 0, that
()' + = 0', and + 0' = 0.
From the commutative law, 0'.
2 The right side of (1-5) should read (ah) -f (ar). However, by conven-
tion, we agree to omit the parentheses when all multiplications are to be
performed before any addition.
4 Introductory Topics Sec. 12
Negative of a Number. It is assumed that for every real number
a there exists a corresponding number, called the negative of a and
designated by a, such that
(1 7) a + (- a) = 0.
For example,
1 + (- 1) =0, (- 2) + 2 =0.
That each number has but one negative may be shown in the
following way: Let x be another negative of a, so that a + x 0.
Then
- a = (- n) +0 = (- <i) + (a + x).
By associativity,
- a ~ ((- o) + n) + x,
or
a + x .r.
In particular, the negative of zero is
- - - + = 0.
The Unit. It is assumed that there is a special number called the
unit and denoted by 1, such that, for every real number a,
(1 8) a 1 = a.
There cannot be a second unit 1'. If there were, we could say that
i . r = i, r- 1 - r,
whence 1 = 1'.
Reciprocal of a Number. It is assumed that for every number a
which is not 0, there is an associated number - > called the rccip-
a
rocal of a, such that
(1-9) a '7 = L
The reader may verify the fact that there is only one reciprocal
of each number. Thus, if ,r is another reciprocal of a, that is, if
a x 1, then x ~ -
a
It is important to note the restriction a ^ in the definition of
the reciprocal. In the next section we shall see why this restriction
is needed.
Subtraction. The difference a 6, of any real numbers a and
6, is defined by
(1 10) o - b =a + (- 6).
Sec. 14 /nfroc/ucfory Topics 5
The operation indicated by the si#n minus which produces for any
two real numbers a and b the real number a b is called
subtraction.
Division. The quotient a f b or . or a -r- b of any real numbers a
t)
and &, where b ^ 0, is defined by
'
The operation associating with real numbers a and b (b =/- 0) their
quotient is called division.
It should be noted that subtraction and division are subordinate
to addition and multiplication, in that they are defined in terms of
these latter. The difference a b is that number .r for which
b + .r - a. Also, tho quotient a b is that number y for which
b y ~ a. It should be noted that + (-) = for every
number a, and that a 'a a (I/ a) 1 for every number a ~f 0.
1-3. OPERATIONS WITH ZERO
It has already been noted that the special number has the prop-
erty (t + - a for every real number a. In particular, we may let
a = to obtain
+ = 0.
It has already been noted that = 0, so that = a 4- = a
for every real number a.
Next, we prove that for every real number a,
(1-12) a () -r 0.
Let x a 0. Then, by the distributive law,
.r = a = a (0 + 0) = a - + a - x + x.
If we add -.r, we obtain
= ;r + (- .r) = (.r + .r) + (- .r) = x + (x + (- x)) = x + = x.
Since a: = a 0, (1-12) is established.
From this last result, it follows that, for b - 0,
013) H
1-4. RECIPROCALS
It was noted that every non-zero number has a reciprocal. We
can now see why cannot have a reciprocal. If has a reciprocal x,
then x = 1. Since it has been shown that x = 0, we would have
6 Introductory Topics Sec. 1-4
to conclude that = 1. However, if = 1 is allowed, then for every
number a we have
Hence, would be the only number in the number system. This
situation obviously should be ruled out. Therefore, cannot have a
reciprocal. Moreover, since a/6 = a (1/6), the quotient a/6 is not
defined when 6 = 0.
The reciprocal of the product of two non-zero numbers can be
expressed in another way :
n 1^ 111
(1-J4) - r ---,*
^ ' a b a b
provided that neither a nor 6 is 0. To prove this result, we begin
with
11 7 1 I * , i ,
-.-.&&=- a 'T-O I 1 =1.
a b a b
We then multiply by r to obtain
L_ -1 ! i -!
. i
_
r , -. . ,
a b a b a b a /; a b a b
In the preceding proof, free use has been made of the commutative
and associative laws.
Finally, if a ^ 0, the reciprocal of the reciprocal of a is a itself.
Thus,
(1-15) ^ = a.
Since
^- (!/) = !,
multiplication by a gives
a - 1 a = -T- (I/a) a = -7- 1 = . -
I/a I/a I/a
1-5. THE REAL-NUMBER SCALE
Real numbers may be represented by points on a straight line.
On such a line select an arbitrary point as origin and lay off
equal unit distances in both directions, as shown in Fig. 1-1. (The
i i i i i i i i i i i i i i i
-7-6-5-4-3-2-10 I 2 3 4 5 6 7
Fiu. 1-1
unit segment may have any length whatsoever.) Label the points
thus far specified as indicated : is the origin, 1 is the first point to
Sec. 16 Introductory Topics 7
the right, 2 is the second point to the right, 1 is the first point to
the left, and so on. Rational numbers that are not integers cor-
respond to certain other points in a natural way. For example, 1/2
corresponds to the midpoint of the segment joining points labeled
and 1; and 7/3 represents the point one-third of the dis-
tance from the point 2 to the point 3. It is a basic assumption
concerning the real numbers that every point corresponds to a
unique real number, and that every real number corresponds to
exactly one point. The full significance of this assumption cannot
be developed in an elementary text.
One observation of importance can be made at this time. The
non-zero real numbers are divided into two classes. One class con-
sist of numbers representing points to the "right" of 0, and the
other consists of numbers representing points to the "left" of 0.
The first class consists of positive numbers, and the second of
negative numbers. The number may be considered as constituting
a third class. It is understood that no two of the three classes zero,
positive numbers, and negative numbers have any numbers in
common. Thus a number cannot be both positive and zero, both
positive and negative, or both negative and zero. The specific desig-
nation of any negative number will include an explicit sign , which
is prefixed. (This convention, however, does not exclude the possi-
bility of allowing a general symbol, such as x, to stand for a nega-
tive number.) Positive numbers do not require such a sign,
although frequently the sign f is used.
It is to be assumed that the sum of two positive numbers is
positive, as is also the product of two positive numbers.
1-6. RULES OF SIGNS
To operate effectively with real numbers, a knowledge of the
rules of signs and of properties of negative numbers is essential.
In each of the following relationships, a and b are any two real
numbers, except that the denominator of a fraction may not be zero.
(1-16) - (- a} = a.
(1-17) - (a + 6) = - o - 6.
(1-18) - (a - 6) = - a + 6.
(1-19) (- a)b = - (6); in particular, (- 1)6 = - 6.
(1-20) (-)(- 6) =<*b.
(1-21) -L = - J
8 Introductory Topics Sec. 1-6
a a a
-b b b
a _ a
^T ~ 6'
(1-22)
Proo/s o/ (i-i
(1-16) - (- a) = - (- a) = a + (- a) + (- (- a)) = o + = a.
(1-17) - (a + b) = + - (a + V) = - a + a + (- &) + b - (a + 6)
= - a - 6 + (a + 6) - (a + 6)
= -a-6 + = -a-6 by (1-6), (1-7).
(1-18) - (a - 6) = - (a + (- 6)) = - a - (- 6) = - a + 6
by (1-17), (1-16).
(1-19) Since (- a) -6 + 0- 6 = (- o + a) 6 = 0-6 = by (1-12),
(- a ) b = (- a) b + a b - (a b) = - (a 6) = - (a b).
(1-20) , (- a) (- 6) = - ((- a) - 6) = - (- (a - 6))
= ab by (1-19), (1-16).
- by(1 - 20) -
/i oo\ 1
(I- 22 ) = <" =
a , N 1 / 1\ a i /i in\
= ( ~ a) '6 = ~ ( a- ft) = ~ 6 by (1 ' 19) *
d-23) ^| = - (- |) = | by (1-22), (1-16).
, It has already been observed that non-zero numbers are divided
into two classes, namely, positive and negative. It is assumed that
if # is positive, then -a is negative ; and that if a is negative, then
a is positive. All calculations involving negative numbers can be
made by performing calculations with positive numbers and apply-
ing one or more relationships just given. It follows from (1-17),
for example, that the sum of two negative numbers is negative, and
is equal to the negative of the sum of the negatives of the given
numbers. Also, from (1-20) it follows that the product of two
negative numbers is positive, and is equal to the product of the
negatives of the given numbers. By (1-19) the product of a posi-
tive number and a negative number is negative.
Sec. 1-7 Introductory Topics 9
1-7. FUNDAMENTAL OPERATIONS ON FRACTIONS
A further study of the algebra of real numbers leads us to the
consideration of the fundamental operations as applied to fractions.
By definition, a fraction is the quotient obtained by dividing
one number a by another number 6, where b is not zero. We
call a the numerator and b the denominator; and we generally
write the fraction a/6, read "a over b" or "a divided by b."
We shall list the following basic relationships for applying the
four operations to fractions. In them a, b, c, d are any real num-
bers, except that no factor in the denominator of a fraction may
be zero.
tt-24) ac _ a
(1 24) bc~b
(1-25) + U*.
(1-27)
(1-28)
a .b __ ad + be
c d ~~ cd
c c c
a _ 6 _ ad be
c d ~~ cd
a c ac
b I d b c be
A special case of (1-30) is
/ c _ d
/ d ~ c '
which states that the reciprocal of a fraction is found by inverting
the fraction. Also, by (1-30), dividing by a fraction is equivalent
to multiplying by its reciprocal.
Proofs of (1-24) to (1-30) :
(1-29) Since ~ ^ = by (1-14),
ac 1 1 , , 1 1 lac
7" * ~5 = a 7" * C * "1 "" (& * w * IT * 3 = a C T"~5 == T~5
b d b d b d bd bd
U-24) 2 = J. = J.1=J by (1-29).
6c6c66 ^ \ /
/^ rp\ a,6 l.,l /"'i\l a + 6 v/ie\
(1-25) - + - = a .- + 6 .- = (o + 6) ,- = ^t- .by (1-5).
10 fnfrocfucfory Topics Sec. 1-7
<'- 26 > f+l-ffz + ra-^ *->. *>
/ rt *x # 6 a . ( 6) 1 , f , x 1 / , N 1
(1-27) --- = ~ + - - - = a - + ( 6) - = (a - 6) -
v ' c c c c c v ' c v ' c
= by (1-22), (1-5).
a & _ a (- 6) _ a d (- b) c
-____ __ _ _ +
t"\ On\ t* / Vx U/ J. Cfr W/U/ fl(* ttCt
c
d
c d
c-d
~~ cd
a ic
a
1 a
ad
b/d
b
c , c
d b 'd
'd'
=
a
b
. d
c
erf be
by (1-29), (1-24).
Example 1-1. State which of the fundamental assumptions are employed in
each of the following equations:
o) 3 +9 + (-5) =3 + (-5) +9.
6) 11 + (6 + 3) + 7 = (11 + 6) + (3 + 7).
c) (2 3) 5 = 2 (3 5).
d) 5(3 + 4) = 5 3 + 5 4.
Solution:
d) The associative and commutative laws for addition.
6) The associative law for addition.
c) The associative law for multiplication.
d) The distributive law.
Example 1-2. In each of the following, perform the indicated operation:
a) 2+3. b) (-2) -f (-3). c) 5 + (-3).
d) (+5) -(+3). e) (-7) -(+6). /) (-7) -(-6).
Solution: Each case can be treated as an addition.
a) 2 + 3 = 5.
6) (-2) +(-3) = -5.
c) 5 + (-3) =2.
d) (+5) -(+3) = +5 + (-3) = +2.
e} (- 7) - (+ 6) = - 7 + (- 6) = - 13.
/) (-7) -(-6) = -7 + <+6) = -1.
Example 1-3. By using the fundamental assumptions and rules for operations^
and transforming the left side into the right side, justify the equation
(a + 6) - (c - d) = (6 - c) + (a + d),
Sec. 1-7 Introductory Topics 1 1
Solution: By (1-18),
(a + 6) - (c - d) = (a + 6) - c + A
By (1-1) and (1-2),
(a + 6) - c + d = (b - c) + (a + d).
Example 1-4. By using the fundamental assumptions and rules of operations,
justify the equation
b ac _ be
a b c "~ 6 c
Solution: By (1-29) and (1-24),
6 ac _ a (be) __ 6c
a 6 c "~ a (6 c) ~~ 6 c
EXERCISE 1-1
1. Identify the fundamental law or laws that justify each of the following equations:
a. x + ?/=?/ + x. b. rs = sr.
c. 2(3 5) = (2 3)5. d. 5(a + b) - 5a + 56.
e. (a -f 2) (6 - 3) = (6 - 3) (a + 2). f. (a + b)c = c(a + b) = ca + cb.
2. Find the value of each of the following:
a. (-3) +(+5). b. (-5) +(-3). c. (-1) -(-2).
d. (+7) -(+2). e. (-8) -(-9). f. 0-(-2).
g. (- 5) - 0. h. 15 + (- 3). i. (- 7) - (- 5).
j. (-5) - (+5). k. + (-3) -(+4). 1. (+32) -(-23) +(-45).
3. Evaluate each of the following:
a. (+2) (-3). b. (-3) (-5).
c. (-7) (+5). d. (-5) (-9).
e. 2(-5). f. (-7)0.
g. (-!)(- 2) +(-3)(0). h. (-4) (-5) -(_2)(-l).
i. (+2) (-3) -(+7) (-5).
4. Determine the negative of each of the following :
a. 5. b. - 3. c. 0. d. 2x.
e. 2/3. f. 2-3. g. 2a - 36. h. - (x - y).
i- - [(a) (- &)] J- 3# + 2. k. x - 3. 1. a + 0.
5. Find the reciprocal of each of the following: *
a. ], b. 2/3. c. 3 +|- d. 2 +
* O t
e. 1.02. f. a +6. g. -- 1 ' h.
o a
i _L_. i ,_ 2? - k l ' 5
! r |^ j^ &J t A*
x + y * * * - r - 0.1 " 2 - 0.3*
12 Introductory Topics Sec. 1-7
6* Prove each of the following equations by using rules for signs and for operations
with fractions:
a. a ( - 6) + a ( - c) = - a (b -t- c). b, - (ac - ad) = a[d + ( - c)].
c. - [6 - (a - c)] = (a - 6) - c. d. 6 / -- h c = a.
/ rt ~~* C
<J&C OC jr/
o- f - <*-< -
1-8. ORDER RELATIONS FOR REAL NUMBERS
We shall use the notation a > to express the fact that a is a
positive number, and the notation a < to indicate that a is nega-
tive. The symbol > means is greater than, and < means is less
than. These symbols are called order symbols.
Assume that a and b are any two given numbers. If a - b > 0,
we shall write a > 6, or b < a, and shall read "a is greater than b,"
or "& is less than a." As can be easily seen, a > b means that a lies
to the right of 6 on the real-number line. When a > b, that is, when
a b > 0, then & a, which equals (a 6) by (1-18), is nega-
tive; and conversely. Hence, a>&(or&<a) if and only if
b - a < 0.
The student is familiar with the symbol - (for equality), which
is used to indicate that two quantities are the same. Thus a = b
means that the two symbols a and b represent the same mathe-
matical object. For example, 6-3*2.
If a and 6 are two distinct numbers on the scale, we say "a is
different from &" or "a does not equal &," and we write symbolically
a^b. The symbol = means does not equal and is called the
inequality symbol. \
In general, the oblique line or vertical line through any symbol
will form a new symbol which is the negation of the original one.
Thus, a < b means "a is not less than b." In other words, a = b or
a > b (by Property 1 below). For example, 5 < 3.
Sometimes we shall find it convenient to combine the symbols
< and = or > and =. We write ^ to mean is less than or equal to,
and we write ^ to mean is greater than or equal to.
We thus have order relations on pairs of real numbers, defined
by either of the following equivalent statements :
a > b (or b < a) if and only if a b is positive;
a > b (or 6 < a) if and only if b a is negative.
The system of real numbers is then said to be ordered by the rela-
tion > (or the relation <) . Assertions of the type a < b or a > b are
See. 1-8 Introductory Topics 13
called inequalities. The ordering of the real numbers has the fol-
lowing properties.
Property 1. For every pair of real numbers, a and 6, one and
only one of the following relationships holds:
a = 6, or a < 6, or a > b.
Proof of Property 1: If a = 6, the statement is certainly true.
Saying that a = b is equivalent to saying that a b = 0, so that
a 6 is either positive or negative. Thus, if a 6 is positive, we
have a > b. If, however, a b is neither positive nor zero, then it
is negative, and a < b. If two of the three possibilities occurred
together, we should have, say, a = b and a > b, or a > b and a < b.
Thus, a b would be both zero and positive, or both positive and
negative. Since no overlapping may occur among the three classes
of numbers, we are thus led to a contradiction.
Property 2. For any real numbers a, 6, c, it is true that
if a < b and b < c, then a < c.
Proof of Property 2: If a< b and 6 < c, then both 6 a and
c b are positive. Let us write c a as (c 6) 4- (b a). We
have assumed that the sum of two positive numbers is positive.
Since c b and 6 a are positive by assumption, their sum, which
is c a, is also positive. Hence, c > a, or a < c.
Property 3. For any real numbers a, b, c, it is true that
if a > by then a + c > b + c.
Proof of Property 3: By definition, a> b means that a 6 is
positive. But, by (1-17),
(a + c) - (b + c) = a + c + (-b) + (-c) = a - 6.
It therefore follows that (a + c) - (6 + c) is positive, and that
a + c > b + c.
Property 4. For any real numbers a, 6, c, it is true that
if a > b and c > 0, then ac > be.
Proof of Property 4: We have assumed that the product of two
positive numbers is positive. Since both a 6 and c are positive, it
follows that their product is also positive. But (a 6)c = ac be.
Therefore, ac be is positive, and ac > be.
Property 5. For any real numbers a, 6, c, it is true that
if a > b and c < 0, then ac < be.
14 Introductory Topics Sec. 1-8
Proof of Property 5: We have noted that the product of two
numbers with unlike signs is negative. Here a - b is positive, but
c is negative. Since (a 6) c = ac - 6c, and (a - 6) c is negative, it
follows that ac be < 0, or that ac < be.
According to Property 3, the order symbol in an inequality is not
changed if the same number is added to or subtracted from both
sides. It therefore follows that a term on one side of an inequality
may be transposed to the other side with its sign changed. For
example, if a 6 > c, then a > c + b.
According to Property 5, the order symbol in an inequality is
reversed if both sides are multiplied or divided by the same nega-
tive number.
1-9. ABSOLUTE VALUE
As a consequence of the properties of the ordering of real num-
bers, there can be associated with each number a certain non-
negative number called its absolute value. For any real number a,
we define the absolute value of a, denoted by |a|, as follows:
| a | = a, if a ^ 0, and | a \ = a, if a < 0.
Thus, | 3 | = 3, since 3 > 0; also |- 3 | = - (- 3) = 3, since - 3 < 0.
1-10. INEQUALITIES INVOLVING ABSOLUTE VALUES
We shall now consider some inequalities involving absolute
values. If we let the number x be represented by a point P on a
number scale, then \x\ is the numerical distance between P and the
origin. If we let a be a positive number, then \x\ < a means that
the point P is less than a units from the origin; that is, x lies
between a and a. We can write this in the form a < x and x < a,
or more briefly in the form a < x < a. Therefore, the statements
\x\ < a and a < x < a mean exactly the same thing.
A more general inequality which often occurs is |# 6| < a,
where a > 0. This is equivalent to a < x 6 < a. If 6 is added
to each term, we may write 6 a< x < b + a. Hence, the state-
ments \x b\<a and 6 a<x<b + a mean exactly the same
thing.
For example, \x~ 3|<2 may be written 2<x 3<2 and
means that the distance between x and 3 is less than 2. To solve
this inequality for x t we add 3 to each term of the inequality,
obtaining 1< x < 5.
The following illustrative examples may help to give a better
understanding of the processes involved in the solution of the prob-
lems in Exercise 1-2.
Sec. 1-10 Introductory Topics 15
Example 1-5. Arrange the following numbers in increasing order:
2, - 3.5, O,TT, 3.14, |-5 |.
Solution: Since TT is approximately 3.1416, the desired order is as follows:
- 3.5, 0, 2, 3.14, TT, |-5|.
Example 1-6. Insert the proper inequality sign (order symbol) between the
following numbers:
- 2 and |-2 |.
Solution: Since | - 2 | = 2, and since - 2 < 2, we have the inequality - 2
< |-2 |, or |-2 | > -2.
Example 1-7. Find integers a and 6 such that a < \/2 < b.
Solution: Since \/2 may be represented approximately by 1.414, the values
a = 1 and 6=2 satisfy the inequalities. Thus, 1 < \/2 < 2. Any other pair of
integers a and 6 such that a ^ 1 and 6^2 would also satisfy the inequalities,
Example 1-8. Express the inequality | x \ < 3 without using the absolute-value
symbol.
Solution: We know that the statements | x \ < a and a < x < a mean exactly
the same thing. Here a is the positive number 3, and | x \ < 3 means that the
point represented by x is less than 3 units from the origin; that is, x is between 3
and 3. The inequality may be written 3 < x < 3.
Example 1-9. Explain the meaning of the inequality | x 2 | < 1 and write it
without using the absolute- value symbol.
Solution: The inequality | x b \ < a is equivalent to a < x b < a. Hence
| x 2 | < 1 may be written - 1 < x - 2 < 1. If we add 2 to each term of the
inequalities, we obtain 1 < x < 3.
EXERCISE 1-2
1. Arrange the numbers in each of the following sets in increasing order:
a. - 3, 0, 4, - 2, 5. b. - 6, - 8, 2, 0, 1/2, - 3/4.
c. - 2, 10, - 1, - 1/3, - 4. d. - 10, 9, 4, - 3, 3/8, - 6/5.
e. 3, - 2, 1, V3, - 3/2. f . 1.4. 0, - 2, V, I - 3 |.
2. Insert the proper order symbol between the two numbers in each of the
following: ,,
a. 3 and 1/3. b. - 3 and | - 3 |. c. \/2 and 1.414.
d. - 3 and - 2. f e. 22/7 and w. t. 1/8 and 1/6.
3. Examine each of the following inequalities, and determine whether or not it is
true.
a. - 5 > - 3. b. - 3 -f 2 < 0. . c. | - 3 | > - 3.
d. TT > 22/7. e. | - 2 | < | 2 |. f. | 3 - 7 | > | 5 - 2 |,
16 Introductory Topics Sac. 1-10
4. Find the value of each of the following:
a.|+2|-|-2|. b.|-3| + |+3|. c.|+4|-|-4|.
d. |-7| + |-5|-|+5|. e. | 12 - 4 | - | - 6 |. f. |5-3|+|3|-|2|.
g. (-18)* 3. h. |-9|*|4|. 1.0*14.
j. 0|4|-|-5|.
5. Express each of the following inequalities without using the absolute-value
symbol:
a. I x I < 1. b.
< 1. c. \x I ^ a.
2
d. | 2x | < 4. e. | x - 1 | < 3. f. | s - 1/2 | < 3/2.
6. In each of the following, find a pair of integers, a and 6, such that the given
inequalities are satisfied:
a. a < 5 < b. b. a < - 3 < b. e. a < < b.
d, a < TT < 6. e. a < \/3 < &. f . a < 1 1 - 2 | < 6
7. If a ^ 3, place the proper order symbol between a + 7 and 10.
8. If a *z 5, what can be said about the value of 3a 2?
1-11. POSITIVE INTEGRAL EXPONENTS
If two or more equal quantities are multiplied by one another,
the product of the equal factors is called a power of the repeated
factor. Thus 5 2 , read "5 squared," means 5 5 ; 5 3 , read "5 cubed/'
means 5-5-5. In general, a n means the product of n factors each
equal to a. We call a the base and n the exponent of the power. It
follows from the associative law that
a 2 a 3 = (a a) (a a a) = a a a a a = a 5 = a 2 * 3 .
Also, if a ^ 0,
a a a a a a
a 3 a a a
= a a = a
= "2 - ,,5-3
These and similar results suggest the following laws of expo-
nents. In them m and n are positive integers. The proofs of these
laws are reserved for a later chapter.
Law of Multiplication. To multiply two powers of the same base,
add the exponents :
(1-31) a m a n = a m+n .
Law of Division, To divide one power of a given base by another
power of the same base, subtract the exponents :
(1-32) ' ^ = a"*-*, if a ^ 0, m > n.
Sec. 1-12 Introductory Topics 17
Law for a Power of a Power. To raise a power of a given base to a
power, multiply the exponents :
(1-33) (a m ) n = a mn .
For example, (a 3 ) 2 = a 3 a 3 = a 3 ' 2 = a 6 .
Law for a Power of a Product. To obtain a power of a product,
raise each factor of the product to the given power :
(1-34) (aft)* = a n b.
Thus, (3a 3 ) 2 = 3 2 (a 3 ) 2 = 3 2 a 6 = 9a 6 .
Law for a Power of a Quotient. To obtain a power of a quotient,
raise the numerator and the denominator to the given power :
(n\ H n n
I) =, if 6*0.
Thus if 6*0
inus, n o * o,
1-12. ALGEBRAIC EXPRESSIONS
An algebraic expression is formed by combining numbers by
means of the fundamental operations of algebra. The distinct parts
of the expression connected by plus and minus signs are called
terms. The terms of the expression 3x 2 -5xy 2 +7z are 3# 2 , -5xy 2 ,
and 7z. Here the numbers 3, 5, and 7 are called numerical coffi-
cients, or just coefficients; x 2 , xy 2 , and z are called the literal parts.
An expression containing one or more terms is called a multi-
nomial. A multinomial consisting of one term is a monomial. A
binomial is a multinomial consisting of two terms, and a trinomial
is a multinomial with three terms. A polynomial is a multinomial
whose terms are of the form ax m y n z p , where m, n, p, are
positive integers and a is a numerical coefficient, and where one or
more of the factors x m , y", z p , may be absent. Thus, 7, 5# 4 , and
3xy + 2 are polynomials, while x + - is not.
y
The degree of a term of a polynomial is the sum of all the expo-
nents in its literal part. For example, the degree of 3# 2 is 2, the
degree of 5xy 2 is 3, and the degree of 7z is 1, because the sums of
the exponents are, respectively, 2, 3, and 1.
The degree of a polynomial is the degree of its highest-degree
term. Thus, in the trinomial 3x 2 -5xy* + 7z, the third-degree
term, -5xy 2 , is its highest-degree term? Therefore, 3# 2 - 5xy* + 7z
is a polynomial of the third degree.
18 Introductory Topics Sec. 112
By a polynomial in x of degree n we mean an expression of the
form
a x n + aix n ~ l + h a n ,
where the coefficients a , &i, * , a n are numerical coefficients,
a T^ 0, and w is a positive integer. If n = 0, we agree that the poly-
nomial reduces to a number a , which is not 0, and that the degree
is zero. The number is regarded as a polynomial also, but as one
having no degree.
If the typical polynomial just given has degree n, the coefficient
a is called the leading coefficient. If its leading coefficient is 1, a
polynomial is called monic. Thus, x* 2x 2 + 5x + 1 is a monic poly-
nomial of degree 3.
1-13. EQUATIONS AND IDENTITIES
An equation is a statement of equality between two numbers or
algebraic expressions. The two expressions are called members, or
sides, of the equation. Equations are of two kinds, namely,
conditional equations and identities. A conditional equation, or
simply an equation, may be true only for certain values (possibly
none at all) of the literal quantities appearing. An identity is true
for all numerical values that can be substituted for the literal
quantities.
Illustrations of equations are
3x - 5 = x + 1
and
x 2 - 5x + 4 = 0.
The first one is true only if x = 3, and the second is true only if
x = 1 or x = 4.
Illustrations of identities are
3(s - 2) = 3x - 6
and
x 2 - 5x + 4 = (x - 1) (x - 4).
Each of these equations is true for all values of x.
1-14. SYMBOLS OF GROUPING
Parentheses ( ) and other symbols of grouping which have the
same meaning as parentheses, namely, brackets [ ], braces { }, and
the vinculum , are used to associate two or more terms which are
to be combined to form a single quantity. The word "parentheses"
Sec. 1-14 Introductory Topics 19
is often used to indicate any or all of these symbols of grouping.
Removal of the symbols of grouping is accomplished by applying
the laws of algebra, such as the laws of signs and the distributive
law. The following examples illustrate the procedure.
Example 1-10. Remove parentheses from - (2x - 3).
Solution: The steps may be indicated as follows:
- (2x - 3) = ( - 1) (2x - 3)
= (-l)(2x) +(-!)(- 3)
= - 2x + 3.
Example 1-1 1. Remove symbols of grouping from 8x 2[5y + 3 (x y)]
[2y x - 3y} and collect terms.
Solution: One way of obtaining the desired result follows:
8x - 2[5y + 3(z - y)] - {2y - x - Zy]
= Sx - 2[5y +3x - 3y] - {2y - x + 3y}
= Sx - 2[2y + 3x] - {5y - x}
= 8x - 4y - 6x - 5y + x
The basic rules for enclosing a group of terms in parentheses may
be stated as follows :
To write a given expression in parentheses preceded by a plus
sign, write the terms as they are given, enclose them in parentheses,
and write + in front of the parentheses. Thus,
To write a given expression in parentheses preceded by a minus
sign, change the sign of each term, write the resulting terms in
parentheses, and write in front of the parentheses. Thus,
a - & = - ( - + 6) = - (6 - a).
The first rule is obvious, and the second follows from the rule of
signs (1-18).
Example 1-12.
a) Enclose the last two terms of 2 4- 3x y within parentheses preceded by a
plus sign.
6) Enclose these terms within parentheses preceded by a minus sign.
Solution:
a) Since the sign before the parentheses is to be +, we enclose 3x - y in its
given form within the parentheses preceded by a plus sign. In this case the ex*
pression becomes 2 -f (3x - y).
20 Introductory Topics Sec. 1-14
6) Since the sign before the parentheses is to be , we change the sign of each
term of 3x - y and enclose 3x + y within the parentheses preceded by a minus
sign. Then the expression becomes 2 ( 3# + y).
1-15. ORDER OF FUNDAMENTAL OPERATIONS
Parentheses and other symbols of grouping are useful in indicat-
ing which operation is to be performed first. We have used them
in this way from the outset. In order to avoid using them unneces-
sarily, as has been already pointed out, the convention is adopted
to perform all multiplications first and then the additions (or
subtractions). If two or more of these symbols of grouping are
used in the same expression, we usually (though not necessarily)
remove the innermost pair of symbols first.
Illustrations in which symbols of grouping are removed follow :
a) 4 - (6/2) =4-3 = 1.
6) (4 - 6)/2 = (- 2)/2 = - 1.
c) 6 + [15/(3 5)] = 6 + (15/15) =6 + 1=7.
d) 6 + (15/3) '5 = 6 + 5-5 = 6 + 25 = 31.
EXERCISE 1-3
In each problem in the group from 1 to 36, perform the indicated operation or
operations.
1. 2 5 . 2. 3 4 . 3. (- I) 3 . 4. 5 3 .
5. 10'. 6. (- 2) 4 . 7. (- 6) 3 . 8. - 10 6 .
9. 7*. 10. (-4) 8 . 11. (|) 4 - 12
()' "' * (-!)' *>
17. i(4 3 ) 18. f(3 4 ) 19. a(5 8 ). 20. a 2 6(-
J o
21. a 3 a 4 . 22. (a 3 ) 4 . - 23. (a&) 4 . 24.
25. ( 2 6 3 ) 2 . 26. a a 2 a 3 . 27. a 2 a 4 a 7 , 28.
. 34. &Y- ttpV. 36. (W-
a 8 V4/ w/ \a 8 /
In each problem in the group from 37 to 48, remove symbols of grouping and
simplify.
37. 3 - (b - 2). 38. x + (y - z).
39. 4[a + (6 - a)]. 40. (a + 26) - (3a - b).
41. a - t + (2a - 36). 42. a - [3 + (2a - 4)].
43. (a - 26) - (3a + 6). 44. W - [3x + (2y - z 2 )].
Sec. 1-16 Introductory Topics 21
45. a* 2 - 2bxy + [(6y - 2cx*) - (axy + i/ 2 )].
46. 3n6 { 4ac -f [ab 2ac + ab] 3a6 } .
47. 2a - [36 -f 4c - (3a - 6 + a~^~6 - (3a 4- 2c) } - 4c + a].
48. - { -[ -(a -b -c) -a + (6 -c)]J.
In each problem from 49 to 60, enclose the last two terms in parentheses. First
use a plus sign before the parentheses, and then use a minus sign.
49. a + b + c. 50. a 2 - 2ab + 6*. 51. a 2 - 6 2 + c 2 .
52. x +y -1. 53. 2a + 6 - 3c. 54. 3x - 4y + 20.
55. z 2 - ?/ 2 - z 2 . 56. - x 2 - ?/ 2 - z 2 . 57. - a 3 6 + ab + & 2 -
58. ax 2 - 2ax?/ + y*. 59. 2z - 3y - 4z. 60. - x 2 - 3 + 1.
In each of the following problems, evaluate the given expression.
61. 16 - (6 - 2). 62. 16-6-2. 63. (- 3) (-4) - (4) (- 2).
64. 4 - (6 - 7). 65. 4 6 - 7. 66. (4 6) - 7.
67. (3 3) - (4 2). 68. 3 3 - 4 2. 69. 3 (3 - 4) 2.
70. 3 3 - (4 2). 71. (8/2) + 4. 72. 8/(2 + 4).
73. 8 + (4/2). 74. (8 + 4)/2. 75. 8 + 4(1/2) + 3.
76. (8 + 4)/(2 + 3). 77. 8 + (4/2) + 3. 78. 8 + ((4/2) + 3).
1-16. ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS
Terms, such as 2x 2 y 3 and 5x 2 y B , which have the same literal parts,
are called similar or like terms and may be added or subtracted by
adding or subtracting their coefficients. To illustrate, let us con-
sider an example.
Example 1-13. Add 5zy 2 , 7x*y, - 2zy 2 , - 9x*y,
Solution: Collecting like terms and adding coefficients, we have
(5 - 2)zt/ 2 + (7 - 9)x 2 y + 4z 2 i/ 3 = 3xy - 2x*y
The procedure used in the solution of Example 1-13 follows at
once from the distributive law. For example, the sum of the terms
5xy* and -2xy 2 is obtained as (5 - 2)xy* or Say 2 .
This leads at once to the rule for the addition (or subtraction) of
algebraic expressions. In practice, we usually arrange like terms
in vertical columns, and then we find the sum of each column by
prefixing the sum of the numerical coefficients in the column. The
procedure may be made clearer by means of the following examples.
22 fnfroe/ucfory Topics Sec. 1-16
Example 1-14. Add 2x 2 - 3xy + z, x 2 - 50, 2xy + 3z.
2z 2 - 3z2/ + z
x 2 -5z
2xy -f 3z
3x 2 xy z.
Example 1-15. Subtract 2a 2 - 36 + c 2 from 3a 2 - c 2 .
Solution: 3a 2 - c 2
2a 2 - 36 -f c 2
a 2 + 36 - 2c 2 .
1-17. MULTIPLICATION OF ALGEBRAIC EXPRESSIONS
With the help of the distributive law for multiplication, the
product of two algebraic expressions is found by multiplying each
term of one by each term of the other and combining like terms.
Example 1-16. Multiply 3x 2 - 2xy + y 2 by 2x - 3y.
Solution: 3z 2 - 2xy + y 2
2x - 3y
2xy 2
6z 3 - I3x 2 y H- Sxy 2 - 3y 3 .
1-18. SPECIAL PRODUCTS
The following typical forms of multiplication occur so frequently
that we should learn to recognize them quickly and to obtain the
products without resorting to the general process of multiplication.
They should be learned thoroughly.
(1-5) a(b + c) = ab + ac.
(1-36) (a + b) (a - 6) = a 2 - b 2 .
(1-37) (a + 6) (a 2 - ab + b 2 ) = a 3 + 6 3 .
(1-38) (a - 6) (a 2 + ab + b 2 ) = a 3 - 6 3 .
(1-39) (a + ft) 2 = a 2 + 2a6 + b 2 .
(1-40) (a - b) 2 = a 2 - 2ab + b 2 .
(1-41) (ax + by) (ex + dy) = acx 2 + (ad + bc)xy + bdy 2 .
See. 1-19 Introductory Topics 23
1-19. DIVISION OF ALGEBRAIC EXPRESSIONS
To divide a polynomial by a monomial, divide each term of the
polynomial by the monomial and add the results.
This rule follows immediately from the rule for fractions
expressed by (1-25).
Example 1-17. Divide 6a 2 z 2 - 12a% - 30a 6 z 6 by 15a 3 z 2 .
~ , ,. 6a 2 z 2 12a 4 z 30a 6 x 5
Solution:
15a 3 z 2
_
5a 5x
To divide a polynomial (the dividend) by a polynomial (the
divisor) , arrange both according to descending or ascending powers
of some common literal quantity. Then proceed as follows:
Divide the first term of the dividend by the first term of the
divisor to obtain the first term of the quotient.
Multiply the entire divisor by the first term of the quotient, and
subtract this product from the dividend.
Use the remainder found by this process as a new dividend, and
repeat the process. Continue the work until you obtain a remainder
that is of lower degree in the common literal quantity than the
divisor.
Example 1-18. Divide 6* 3 - 5z 2 + 3x + 1 by 2x - 1.
Solution: 2x - 1 | 6z 3 - 5x 2 + 3x + 1 | 3x 2 - x + 1
6s 2 - 3s 2
- 2x 2 + 3z
- 2x 2 4- x
2x + 1
2x - I
2.
The division can be checked by finding the product of (2x - 1) and (3z 2 - x -f 1)
and adding 2, proving that 6z 3 - 5x 2 -f 3z + 1 = (2s - 1) (3z 2 - x + 1) -f 2.
Any problem in division, in general, may be checked by means df
the relationship ?
dividend = (divisor) (quotient) + remainder.
This equation is an identity ; that is, it is true for all values of the
literal quantities. Indeed, this equation supplies the underlying
meaning of the process of division,
24 Introductory Topics Sec. 1-19
EXERCISE 1-4
In each problem in the group from 1 to 12, add the given expressions which are
separated by commas.
1. - xy, 3xy. 2. 4x 2 y 2 , - 2xy. 3. 3x 2 y 2 f - 2x 2 y*.
4. 6a + 116, - 3a + 26. 5. 4a 2 - 26 2 , a 2 - 46 2 . 6. 3a - 26 + 4c, 4a + 36 - 6c.
7. -3a+6~c, -a-6+c.
8. 3x + 2y - z, x + y - 3z, 4x - 3y + 20.
9. x 2 - 2xy + y 2 , 4xy, - y 2 .
10. 3z 2 - 4j?/ 2 + 2?/ 3 , 4z 2 ?/ - 2x 2 - y 3 , 4xy 2 - 2y 8 .
11. 2x* - 3z + 1, x 2 + 2x - 3, 2z 2 - x s + 4 - 2z.
12. 4az + 3bxy - 4z 2 , 26z - bx 2 + 2azy, 3z - 2y.
In each problem from 13 to 24, subtract the second expression from the first.
13. - xy, 3xy. 14. 4x 2 y 2 , - 2xy.
15. 3z 2 2/ 2 , - 2x 2 2/ 2 . 16. 6a + 115, - 3a + 26.
17. 4a 2 - 26 2 , a 2 - 46 2 . 18. 3a - 26 + 4c, 4a + 3& - 6c.
19. - 3a + 6 - c, - a - 6 + c. 20. z 3 + x 2 + x + 1, 3J 3 - z 2 - -1.
21. x 2 + 2a;y 4- 2/ 2 > 4xy.
22. 3 - 2x + a: 2 - a: 3 + 3z 4 - 2x 5 , 3x 5 + 4x 4 - a; 3 + 2o: 2 - 2x + 1 .
23. o; -
y 4- xy 2 - y 2 , x*y - 4- /
In each problem from 25 to 60, perform the indicated operations.
25. (6*)(-37/). 26. (4s) (-2).
27. (5**y) (- 2sy). 28. (3a) (2b 2 ) (- 4c).
29. (- 3a6) (4k) (- 2a 3 ). 30. (x - y)2a.
31. (3z 4- 2y) (- 2). 32. (6 - 3a) (2).
33. (4xy + 27/ 2 ) (3xy). 34. (4a 2 + 66 2 ) (3a6).
35. (2x + 3y) (x - y). 36. (3a - 6) (a + 26).
37. 4z 2 (z 2 - 2xy - y 2 ). 38. 3xy(2x 2 y + 3y - 4?/ 2 ).
39. (a; 2 + xy - ?/ 2 ) (x - y). 40. (a; 2 - y 2 ) (x + y) (x - y).
41. 4x/(- 2). 42. 3x 2 y/2xy.
43. 4x*y*z/2xy*z. 44. 20z 6 2/ 4 zyiOz 6 2/2 3 .
45. 4a6/(- 4a6). 46. (2xy) 2 /2xy 2 .
'47. (2xy)V2(xy). 48. (a: 2 - 2*y)/(- s).
49. (3xt/ 2 - fay + 9?/ 2 )/(- 3y). 50. (x 2 y 3 - So: 2 ?/ 2 + 4x*y 2 )/x*y*.
51. (x 2 + 6x + 5)/(x + 1). 52. (9x 2 - 60; + l)/(3x - 1).
53. (x 2 - 2/ 2 )/(z - y). ' 54. (x - y) 2 /(x - y).
55. (x* - y)/(x - y). 56. (a; 3 + 3x 2 + 3x + l)/(x + I) 2 .
57. (x -f y 3 )/(x -f y). 58. (x + y) 8 /( + #)
59. (x 3 - y 3 )/(x - y). 60. (x 4 -f x 3 + 3a; 2 + 2x -h l)/(z 2 + 2).
61. Divide x* - y 2 - (x - y) 2 - (x - y) (x y)byx- y.
62. Divide (a + 6) 2 +6(0+6) + 5 by a + 6 + 5.
63. Divide z 4 + 4z 3 + 6z 2 + 4z + 1 first by x + 1 and then by x 2 + 2z + 1.
64. Multiply x 2 - 2x - 3 by x + 4, and divide the result by x + 1.
Sec. 1-20 Introductory Topics 25
65. Divide x 5 + x 4 + 3z 3 - 2z 2 - 3 by x - 1 and add the quotient to the excess of
3z 4 4- 2z 2 - 9z + 7 over 2z 4 + 2x 3 + 7x 2 + 3z + 4.
66. a) Under what conditions will ( - x) n be positive? 6) When will it be negative?
Assume first that x is positive and then that x is negative.
67. For what values of n will ( a) n be equal to a n ?
1-20. FACTORING
Factoring a quantity is the process of finding quantities which,
when multiplied together, yield the given quantity. When a quan-
tity A is expressed as a product B C, B and C are called factors
or divisors of A and are said to divide A. Also, A is called a multiple
of each of JS and C. These concepts are applicable to numbers or
algebraic expressions generally, but are most useful when restricted
to apply to integers or to polynomials. Such restriction will be
adhered to in this book. Thus, when an integer is to be factored,
the factors sought are to be integers. And when a polynomial is to
be factored, the desired factors are to be polynomials.
Let us first review the fundamentals of factoring integers (posi-
tive, negative, or zero) . First, it is clear that every integer n may
be expressed as 1 *n or ( !)( n). Such factorizations are called
trivial. If an integer n, other than 4-1 or 1, has no factorizations
other than trivial ones, then n is called a prime (number). An
integer having a non-trivial factorization is called composite.
Examples of prime integers are 3, 7, and 11; examples of com-
posite integers are 6 and 40.
Let n be a composite integer. Then a non-trivial factorization
m *p exists in which \m\ and \p\ are less than \n\ and greater than
1. If both m and p are primes, then n is expressed as a product of
primes. If not, at least one of m and p, say p, is composite, and so
p r-s. Hence, n = m r s. The process begun may be continued
if any one of m, r, and s is composite, and additional factors may
be found, until the process cannot be continued further, in which
case only prime factors are obtained. We may conclude that the
factoring process must terminate, since at any stage the new fac-
tors introduced are numerically less than their product. The
fundamental theorem of arithmetic guarantees that every com-
posite integer is a product of primes, which "are unique except for
their signs or the order in which they are written. For example,
successive factoring of 156 gives
156 = 39 4 = 13 3 4 = 13 3 2 2,
and the final factors 13, 3, 2, 2 are primes. Another valid factor-
ization of 156 into primes is as follows:
150 = 2-3- (- 13) -(-2).
26 Introductory Topics Sec. 1-20
Let us turn now to polynomials with real coefficients. These may
be polynomials in x, such as x 2 1 ; or polynomials in x and y,
such a3 x 2 + Sxy + 2y 2 ; or, in fact, polynomials in any number of
literal quantities. Every polynomial F has a factorization of the
form
for every non-zero number a. And, of course, the factors I/a and
aF are themselves polynomials. Such factorizations are called
trivial. If a polynomial F has no factorizations other than trivial
ones, then F is called a prime polynomial, or an irreducible poly-
nomial. A polynomial having a non-trivial factorization is called
composite or reducible.
Examples of prime polynomials are Sx + 2, 2x + 2y, and x 2 + S.
Every polynomial in x of the first degree may be shown to be
prime; and certain polynomials of higher (even) degree in x also
are prime. A development of criteria for primeness of polynomials
lies beyond the scope of this book.
Examples of composite polynomials are x 2 4, xy + y 2 and
xz + yz + xu + yu, because each of these has a non-trivial factor-
ization. Thus, we have
x 2 - 4 = (x + 2) (x - 2),
xy + y* = (x + y)y,
xz + yz + xu + yu = (x + y) (z + u).
Let / be a composite polynomial in x with real coefficients, not all
of which are zero. Then it can be shown that there exist poly-
nomials g and h, the degree of each of which is less than that of /,
such that
/ = g. h.
If g and h are primes, then / is a product of primes. If not, we may
proceed to factor (non-trivially) one or both of g and h, and we
can continue this process until primes are obtained. The process
must terminate eventually, since each non-trivial factorization
leads to polynomials of lower degree.
The argument just presented applies only to polynomials in one
literal quantity x. However, the principle may be extended to
apply to polynomials in any number of literal quantities x, y, z,
"- . Thus,
/ = Plp 2 ?,
where p if p 2 , , p n are prime polynomials. The problem of carry-
Sec. 1-21 Introductory Topics 27
ing out actual factorizations for certain types of polynomials is
considered in the next section.
A special class of polynomials deserves particular attention.
This is the class that consists of polynomials in which all the
numerical coefficients are integers. It is possible to prove a factor-
ization theorem such as that just stated, but yielding factors which
are polynomials having only integral coefficients. When the general
theorem can yield prime factors all of which have integral coeffi-
cients, the two theorems give the same result. Otherwise, they
will give different results.
For example, both theorems applied to x 2 4 yield the prime
factorization (x + 2) (x 2) . The polynomial 3# 2 4 has no non-
trivial factors with integral coefficients. However, when other
real coefficients are allowed, we have
3x 2 - 4 = (V3x + 2) (-x/33 - 2),
in which the coefficient \/5 is a perfectly acceptable real number.
Again, when factors with integral coefficients are desired in such a
case as 36# + 24y, we shall agree to remove and factor common
numerical factors, to obtain
36* + 24y = 3 2 2 (Sx + 2y).
Here the prime factors are the numerical primes 3, 2, 2 and the
prime polynomial 3x + 2y.
In what follows, whenever a polynomial has only integral coeffi-
cients, we agree to restrict ourselves to factors of the same kind.
In similar fashion, when the given polynomial has only rational
coefficients, we shall search for factors having only rational
coefficients.
1-21. IMPORTANT TYPE FORMS FOR FACTORING
The equations in Section 1-18 applied "in reverse" are formulas
for factoring. Success in factoring a polynomial therefore depends
on ability to recognize the polynomial as being a particular type of
product and as having factors of a definite form. Verify the follow-
ing type forms by carrying out the indicated multiplications and
learn each form,
Type 1: Common Monomial Factor. From (1-5) we have
(1-42) ab + ac = a(b + c).
Example 1-19. Factor 4x 2 y - toy*.
Solution: x*y - 6xy* = 2xy(2x - 3
28 Introductory Topics Sec. 1-21
Instructions for actually removing the monomial factor in
Example 1-19 may be put this way : Write the common factor, 2xy,
and in parentheses following 2xy write the algebraic sum of the
quotients obtained by dividing successively every term of 4x 2 y
&xy 2 by 2xy, in accordance with the distributive law.
Type 2: Difference of Two Squares. From (1-36) we have
(1-43) a* - b 2 = (a + b) (a - b).
Example 1-20. Factor 9z 2 - 25s/ 2 .
Solution: 9x 2 - 25y 2 = (3z + 5y) (3x - 5y).
Type 3: Sum and Difference of Two Cubes. From (1-37) and
(1-38) we have
(1-44) a 3 + 6 3 = (a + b) (a 2 - ab + b 2 )
and
(1-45) a 3 - 6 3 = (a - b) (a 2 + ab + b 2 ).
Example 1-21. Factor &r 3 - 27y*.
Solution: 8x* - 27 y* = (2x - 3y) [(2x)* + (2x) (Zy) + (3?/) 2 ]
= (2x -3y) (4o: 2 +6xy + 9t/ 2 ).
Type 4: Perfect-Square Trinomials. From (1-39) and (1-40),
(1-46) a 2 + 2ab + b 2 = (a + b) 2
and
(1-47) a 2 - 2a6 + 6 2 = (a - 6) 2 .
Example 1-22. Factor 4z 2 + 12a?y a + 9?/ 4 .
Solution: 4x* + 12z?/ 2 -f 9y* = (2^) 2 4- 2(2z) (Sz/ 2 ) + (3?/ 2 ) 2
= (2x + 32/ 2 ) 2 .
Type 5: General Trinomial. Factorization of the general tri-
nomial may be indicated as follows :
(1-48) Ax 2 + Bxy + Cy 2 = (ax + by) (ex + dy).
If the two factors (ax + by) and (ex + dy) are multiplied
together, the product is found to be
acx 2 + (be + ad)xy + bdy 2 .
By comparing this product with the trinomial
Ax 2 + Bxy + Cy 2 ,
we note that it is necessary to find four numbers a, 6, c, d, such that
ac = A, be + ad = B, and bd = C.
Sec. 1-21 Introductory Topics 29
The following example illustrates a trial-and-error procedure,
which often involves several steps of inspection and testing by
multiplication ; yet it is a method commonly employed in practical
work and is recommended here.
Example 1-23. Factor 6x 2 + llxy - l(ty 2 .
Solution: Here we wish to find a, 6, c, and d which satisfy the identity
(ax 4- by) (ex + dy) = acx* + (ad + 6c)zz/ + fccfr/ 2 = Sx 2 + llxy - 10i/ 2 .
Since ac = 6 and bd = - 10, obviously a and c have like signs, and b and d have
opposite signs. Possible values for a and c are 1, 2, db 3, and 6. Possible
values for 6 and d are db 1, =fc 2, 5, and 10. By trial and error we find the
correct selection to be a = 2, c = 3, 6 = 5, and d = 2. This selection meets the
requirement because
(2x + 5t/) (3s - 2y) = 6z 2
Type 6: Factoring by Grouping. An expression which does not
fall directly into one of the given type forms can sometimes be
reduced to one of these forms by a suitable grouping of terms.
The following examples illustrate the procedure.
Example 1-24. Factor Sax - 56z + Say - 106y.
Solution: First, group within parentheses the terms having a common factor.
Thus,
Sax - 5bx + Say - IGby = (Sax - 5bx) + (Say -
Then, in each group, factor out the common quantity. In this case,
(Sax - 5bx) + (Sax - IGby) = x(3a - 5b) + 2y(3a - 56),
Finally, factor out the quantity common to the terms obtained. (This quantity
will often be a multinomial factor.) The result is
x(3a - 56) + 2y(3a - 56) = (3a - 56) (x + 2y).
An alternate method of grouping would give us the following results :
First,
3ax - 56x + Say - IGby = (3ax + Say) + (- 56x - IGby).
Then,
(3ax + Say) 4- (- 56x - IGby) = 3a(x + 2y) - 56(x + 2y).
Finally,
3a(x + 2y) - 56(x + 2y) = (x + 2y) (3o - 56).
Example 1-25. Factor x 2 - ?/ 2 4- 6y - 92 2 .
Solution: By grouping the last three terms, we may rewrite the given expression
as the difference of two; squares. Thus,
X 2 _ y2 + ey - 9 2 = x 2 - (y* - 6y
= x 2 - (y - 3*) 2
= [x - (y - 3z)} [x + (y -
= (x - y + 3*) (x + y -
30
Introductory Topics See. 1-21
Example 1-26. Factor 4x 4 + 3x 2 ?/ 4 + y*.
Solution: By adding and subtracting x2/ 4 , we may rewrite the given expression
as the difference of two squares. Thus,
4x 4 + 3x 2 j/ 4 + 2/ 8 = 4x 4 + 4* 2 ?y 4 + y* - * 2 2/ 4
(2x 2 + !/ 4 ) 2 - X 2 T/ 4
= [(2x 2 + T/ 4 ) - XT/ 2 ] [(2x 2 + y*) + xy 2 ]
- (2x 2 + ?y 4 - XT/ 2 ) (2x 2 + y* + XT/ 2 ).
EXERCISE 1-5
Factor each of the following:
1 3x + Qy 2. 6x 2 + 4x?y. 3. 4x + 14.
4*. 2a - ab. 5. Sax + 2a. 6. 4ax' - 2/a 2 x3.
7. - ax + 2cx - *. 8. 2at - 36t 2 + . 9. ax - 2ay + 3a.
10. 3a6 + 9ac - 66c. 11. 5t/ 2 + 3y - at/ 2 . 12. x' - 5x 2 t/ + 6*y.
13. x 2 - 16. 14. 2/ 2 - 64. 15. 4x* - 9.
16 16x 4 - 25t/ 4 . 17. 49x 2 - 121. 18. 25a 2 - 166 2 .
19 ' 92/ 2 _ a i y 20. 1 - x 2 ty 2 . 21. a'x 4 - 9a 4 x 2 .
22*. 4a'x 4 - 81. 23. 0.01 - 6 2 . 24. - a'x 2 - ax'.
25. 49x 2 2/ 2 - 144a 2 &'. 26. a 2 - 27. 36* - x.
28 4x 2 - 16x 4 . 29. 2x' + 16. 30. 3x' - Sly*.
31. x 3 ty + z\ 32. a - 2166 8 . 33.
34. a 3 + 6 9 . 35. x - y 12 . 36 -
37. Six" - 3(2?y + 3z) 3 . 38. ^a + 26) 3 + (3c_-t_4d) 8 .__39 i _
.' "" ""41. 216* -l/< 42.
" 12x3 + 3,5. 44. 8x 2 + 2x - 15. 45. x 2 T/ 2 - 18* + 81.
_ 6. 47. x 4 - 8x 2 + 15. 48. x 2 - 11* + 30.
r 2 - 2x - 3. 50. 2x 2 - 3x + 1. 61. x 2 - 4xj/ + 4t/ 2 .
4-1 53 x 2 - x - 12. ' 54. x 2 - 3x - 10.
I 9' W x' 4-4x3+4 57. 1 + 30x3 + 225x.
oa. & -r uu. j + & " * T^ ' ^' _ .
58. x 4 - 10x 2 + 9. -5$. 5x3 + iQx 2 - 40x. 60. a 2 + 5a - /4.
61 2x - 11* - 6. 62. x 2 + 16* + 64. 63. 2** + 5* - 3.
64' 18x 2 + 16* - 25. 65. 6x' - 37* + 6. 66, 4* 2 + 32* + 15.
67 x 2 - 1 2x + 0.36. 68. 40x 2 2/ 2 + 35*yz - 18 2 . 69. a* + 3x + 2ay + 6.
70*. x' + 3x 2 - 7x - 21. 71. 8x' - 12x 2 - 10* + 15.
72. ax + 26x - ex + Say + 66ty - 3cy. 73. 2x + 3y - xz + 6a + xy - 3*.
74. xa6 - xyz - 2aby + 2y*z. 75. x 3 + x 2 - 3x - 3.
1-22. GREATEST COMMON DIVISOR
A common divisor of several polynomials (or integers) is a poly-
nomial (or integer) which divides each of them. For example,
2 and 3 are common divisors of 12 and 18. Also, x and x + y are
common divisors of ^ (a* - y-) and x* (x* + 2xy + y*).
A greatest common divisor (G.C.D.), also known as a highest
common factor (H.C.F.), of two or more polynomials (or integers)
Sec. 1-22 Introductory Topics 31
is a polynomial (or integer) with the following two properties:
It is a common divisor of the given polynomials (or integers) ;
also it is a multiple of every other common divisor of the given
polynomials (or integers).
It follows that, for integers, a G.C.D. is a common divisor of
greatest absolute value. It also follows that, for polynomials, a
G.C.D. is a common divisor of highest degree.
For example, a G.C.D. of 12 and 18 is 4-6 or -6, since 1, 2,
3, and 6 are the only common divisors, and 6 and -6 are those
of maximum absolute value. This example indicates that a set of
non-zero integers will have two greatest common divisors, d and
d, one being positive and the other negative.
For polynomials with real coefficients, if d is a G.C.D., then a d
is also a G.C.D. for every real number a not equal to 0. It follows
that infinitely many greatest common divisors exist. However, for
polynomials with integral coefficients, a G.C.D, should be a poly-
nomial of the same type. Example 1-27, which follows, illustrates
the fact that in this case a G.C.D. is uniquely determined except
for sign.
Since a G.C.D. of given polynomials divides all of them, it must
contain as a factor each of the distinct prime factors occurring as
a common factor of all the given polynomials. Since, however, a
G.C.D. must divide any common factor of the polynomials, it must
contain each of the distinct common primes raised to the highest
common power. Thus, a G.C.D. of x* (x + y) 2 and # 5 (x y) (x + y)
must contain the prime factor x to the third power, that is,
to the highest common power. Similarly, it rftust contain (x + y)
to the first power. Thus, x*(x + y) is a G.C.D., since no further
common prime factors occur.
When no common prime factors occur, a G.C.D. is 1.
Example 1-27. Find a G.C.D. of 3x*y*(x* - 4y 2 ) and 6xy*(x* - 4xy -f 4y*).
Solution: We shall begin by writing each of the expressions as the product of its
prime factors, as follows:
3^3(^2 _ 40.) = ( 3 ) ( X ) ( X ) (y) (y) (y) ( X + 2 y) ( X - 2|0,
and
6xy*(x* - 4xy + y*) = (2) (3) (x) (y) (y) (x - 2y) (x - 2y).
t
The different prime factors are 2, 3, x, y, (x + 2y), and (x 2y), of which only
3, x, y, and x 2y are common to both polynomials. We now form the product of
these common factors, using for each the maximum common power. Hence, a
G.C.D. is
3xy*(x - 2y).
32 Introductory Topics Sec. 1-23
1-23. LEAST COMMON MULTIPLE
A common multiple of two or more polynomials (or integers)
is one containing each of the given ones as a factor. Thus, 36 is
a common multiple of 6 and 9, and x 2 - y 2 is a common multiple of
x y and x + y.
A least common multiple (L.C.M.) of two or more polynomials
(or integers) is a polynomial (or integer) with the following
properties: It is a common multiple of the given polynomials (or
integers) ; also it is a divisor of every other common multiple of
the given polynomials (or integers).
It follows that, for integers, an L.C.M. is a common multiple of
least absolute value. It also follows that, for polynomials, an
L.C.M. is a common multiple of lowest degree.
For example, an L.C.M. of 6 and -8 is 24 or -24, since the only
common multiples are 24, 48, 72, , and 24 and -24 are
those of minimum absolute value. This example indicates that a
set of non-zero integers will have two least common multiples, m
and m, one being positive and the other negative.
For polynomials with real coefficients, if m is an L.C.M., then
a m is also an L.C.M. for every real number a not equal to 0.
It follows that infinitely many least common multiples exist. How-
ever, for polynomials with integral coefficients, an L.C.M. should be
a polynomial of the same type. Example 1-28, which follows, illus-
trates the fact that in this case an L.C.M. is uniquely determined
except for sign.
Since an L.C.M. of given polynomials is a multiple of all of them,
it must contain as a factor each of the distinct prime factors occur-
ring as a factor of any one of the given polynomials. Since, how-
ever, an L.C.M. must be a multiple of any common multiple of the
given polynomials, it must contain each of the various distinct
primes to the highest power occurring anywhere. For example, an
L.C.M. of x*(x + y) 2 and x*(x -y)(x + y) must contain the vari-
ous distinct prime factors, which are x, x + y, and x - y. For x,
the highest power occurring anywhere is the fifth power ; for x + y,
the highest power is the second ; f or x - y, the highest power is the
first. An L.C.M. is therefore x 5 (x + y) 2 (x-y).
Example 1-28. Find an L.C.M. of 4z 2 - 4x, 6z 2 - 6, and 9z 2 - 18x + 9.
Solution: We shall first rewrite each of the expressions in factored form. Thus,
4*2 - 4s = (2) (2) (x) (x - 1) = 2*x(x - 1),
6z* - 6 = (2) (3) (x + 1) (x - 1),
and
90 - 18* + 9 = (3) (3) (x - 1) (x - 1) = 3*(* - I) 2 .
Sec. 1-24 Introductory Topics 33
The distinct prime factors are 2, 3, #, x + 1, and x 1. The greatest powers for
these are 2, 2, 1, 1, and 2, respectively. An L.C.M. is therefore
2 2 3 2 x (x + 1) (x - I) 2 .
EXERCISE 1-6
In each problem from 1 to 12, find a G.C.D. of the given expressions.
1. 4, 14, 36. 2. 9, 21, 33. 3. 4, 7, 39.
4. z + y, x 2 - y 2 . 5. x 3 - y*, x - 37. 6. 3a6, 12a 3 6, 6a 3 6 2 .
7. 9z 3 !/ 2 , 15zV, 21x 6 ?/. 8. x - 3, x 2 - 9, (x - 3) 2 . 9. 4x*y*z, Sxifz*, Ux*y*z 2 .
10. x + 2, x 2 - 4, x 3 + 2z 2 - 4z - 8.
11. x 8 - s a - 42z, z 4 - 49z 2 , z 2 - 36.
12. x 4 + 2z 3 - 3z 2 , 2z 6 - 5z 4 + 3z 3 , z 3 + 3x* - x - 3.
In each problem from 13 to 22, find an L.C.M. of the given expressions.
13. 6, 8, 12. 14. 8, 45, 54.
15. xy, 6xz, 8yz. 16. 4z 2 , 5x 4 , 20z.
17. 4x 2 ^ 4 5 , 9^ 6 2/ 2 2 3 , 6x 4 y02. 18. 2a + 4, a - 3, a 2 - 9.
19. x - 2, x + 2, x 2 - 4. 20. 2x + 8, 3x - 6, x 2 4- 2x - 8.
21. (s - 49) (x 3 - 8), (x - 7) (x + 7) (x - 2) (x 2 - 4), (x - 3) (x - 2).
22. 2x 4 - 22/ 4 , 6x 2 + 12xy + 6s/ 2 , 9x 3 + 92/ 3 .
1-24. REDUCTION OF FRACTIONS
From (1-24) under operations with fractions, it follows that a
fraction a/6, in which 6^0, is not changed if both the numera-
tor and the denominator are multiplied or divided by the same
quantity, provided that the quantity is not zero. That is, if k = 0,
a ka
or
a
6 57*
For example, 2
= 7 > and A =
2
2 ~ 2 2 - 4 ' """ 6 - 6/2 ~ 3 '
The fundamental principles of (1-49) and (1-50) are applied in
reducing a fraction to lowest terms and in changing two or more
fractions with different denominators into equivalent fractions with
a common denominator.
The reduction of a fraction to lowest terms, that is, to a form in
which all common non-zero factors are removed from both the
numerator and the denominator, is accomplished as follows:
First, factor both the numerator and the denominator into
prime factors.
Then, divide both the numerator and the denominator by all
their common factors.
34 Introductory Topics Sec. 1-24
It should be noted that this reduction can also be accomplished
by dividing the numerator and the denominator by their highest
common factor.
The following examples will illustrate the reduction of fractions
to lowest terms.
30
Example 1-29. Reduce -^ to lowest terms.
4<
Solution: Factoring the numerator and the denominator and dividing both by
their common factors, we have
30 _2'3-5 5
42 2 3 7 7 '
The common factors of the numerator and the denominator are 2 and 3.
6xv 2
Example 1-30, Reduce jr-- to lowest terms.
Solution: Factoring into prime factors and dividing out common factors, we have
2 * 3 x y y = 3y
2 x x y x
In this fraction the common factors are 2, x, and y.
_
Example 1-31. Reduce ^-r - ~ - ^-r~- to lowest terms.
6z 2 - 3xy - ISy 2
Solution: We have
2 * 3x(x - 2y) (x + 2y)
6z 2 - 3^ - IS?/ 2 3(s - 2y) (2x + 3y) 2z 4- 3y
It is important to note that in a fraction of the type a ^ y i where
Ct | *C
the numerator and the denominator have a common term, any
attempt to simplify the fraction by cancelling out this common
term can lead only to an absurdity. For, quite obviously, a "^ y
i _i_ ^ *^
does not equal either j] y or - unless a = 1 or 0. For example, if
1 + x x 2-1-3
we attempt one of these simplifications with the fraction ]" >
5453
we reach the obvious contradiction - = - or - = - To avoid this
6564
common error, it is important to remember that only common
factors, not common terms, may be cancelled. One should make
certain that the common quantity is a factor of the entire numera-
tor and of the entire denominator.
1-25. SIGNS ASSOCIATED WITH FRACTIONS
It follows from (1-49) that we can multiply both the numerator
and the denominator of a fraction by 1 without changing the
Sec. 1-25 Introductory Joplcs 35
value of the fraction. However, if just one of these is multiplied
by 1, then the sign of the fraction must be changed in order to
keep its value unaltered. Thus, in effect, a minus sign before a
fraction can be moved to either the numerator or the denominator
without altering the value of the fraction. We have, as previously
stated in (1-22),
- ~~ n n
, d ~" d "~ d
and 1-11 i
x - y x - y (x y) y - x
We see, then, that any two of the three signs associated with a
fraction, namely, the signs of the numerator, the denominator, and
the fraction, can be changed without changing the value of the
fraction. In general, the rules for changing signs in a fraction are
as follows:
Changing the signs in an even number of factors in the numera-
tor or in the denominator, or in both, does not change the sign of
the fraction.
Changing the signs in an odd number of factors in the numerator
or in the denominator, or in both, does change the sign of the
fraction.
2x 2x ( )
Example 1-32. Find the missing quantities in =
7
o ( ) o
7 . . 2x -2x - 2x
solution: = -- - = - -
o o o
Example 1-33. Change the fraction to an equivalent fraction with
denominator x y.
Solution: Since x - y = - (y - x), we make the following changes in signs:
a __ a __ a
or y-x~-(y-x)~~x-y y
a a a
y - x - (y - x) x -y
EXERCISE 1-7
1. Find the missing quantity in each of the following equalities:
a.
d.
g.
4
( )
x 2
( )
3x
-
a
( )
}
7
__i_
I
21
-3
'to
12r 3 ?/ 2 1
" 2
, x
a
-8a
6z 2
2.T
"( )
- 2z
U 3
^ 4-
y
( ) 2x 4- a 4ax -
( ) , x - a (
h2a 2
-x 2
(1 +
*)( )
x -
y
2/ -x
- 6
6 2 -x
!
36
introductory Topics Sec. 1-25
2. Reduce each of the following fractions to lowest terms.
102010 . 8a 2 fe 3 6x*y*
** 350470 ' 12a<6 ' C * 20* V '
f .
96z 3 2/ 7 * 2 ' e * 24*V + 40zV ' 3s 8
a +26 , x*y* - x*y* . 9s 2 - 49
2 ' 4a 2 + 8a6 ' x -y ' K 3z 2 + 13z + 14 '
- 7x - 15 , 6x 2 - 5x - 6 - s - 81
a: 2 -25 4z 2 + lOa; - 24 ' ' x* -f 729 *
_ - 3)2
- 4 - (* - ?/) 2 ' "' a 2 - 462
- -
a; 2 - (a ^- 6)g 4- o& (x + y) (a? - y) (y - z) t
d) - (a: + 2a)2] (a: 2 ~ 6) ' q ' (y* - x 2 ) (z - y)
[(2x
18x 2 4- 3sy - 10?/ 2 9a 2
8y* ' S * 6a 2 - a6 - 6 2
u
-f xy 2 +y 3 * x 2 + x + 2y
1-26. ADDITION AND SUBTRACTION OF FRACTIONS
The sum of two fractions with the same denominator is the
fraction whose numerator is the sum of the numerators and whose
denominator is the given common denominator. That is, in (1-25),
(1-25) -+=^-
For example, x.2y_ x + 2y
3 + 3 3
To add fractions with different denominators, first change the
fractions to equivalent fractions having a common denominator,
and then write the sum of the new numerators over this common
denominator.
Ordinarily, the common denominator which is chosen is a least
common multiple of the given denominators, since this leaves the
fewest possible common factors in the numerator and denominator
of the resultant fractions. However, the same result is obtained,
after compilation of all common factors, no matter what common
denominator is used. The method of finding an L.C.M. was devel-
oped in Section 1-23.
The difference of two fractions, ? > has been defined in
o a
Section 1-2. Thus, by (1-10), when neither 6 nor d is zero,
Sec. 1-26 Introductory Topics 37
Applying (1-22), we have
n r x a c a c
(1-52) T- -- ; = T H -- r- '
v ' b d b d
The following procedure is suggested for adding (or subtracting)
fractions.
1. Find a least common multiple of the given denominators.
2. Change each fraction to an equivalent fraction having the
L.C.M. as the denominator in the following way: For each
fraction, note which factors are in the L.C.M. but not in the
denominator of the given fraction. These factors may be found
by dividing the L.C.M. by the denominator of the given fraction.
Then multiply the numerator and the denominator of the given
fraction by these factors.
3. Write the sum (or difference) of the numerators of the
new fractions found in step 2 over the L.C.M., and reduce the
resulting fraction to lowest terms.
The following examples will indicate a procedure which should
be followed until some skill in working problems has been attained.
357
Example 1-35. Express j - ^ -f Q as a single fraction reduced to lowest terms.
Solution: Step 1. The methods in Section 1-23 give us 36 as the L.C.M. of
the denominators.
Step 2. Divide the L.C.M. successively by the denominators 4, 6, and 9 to get
36/4 = 9, 36/6 = 6, and 36/9 = 4.
Change the given fractions to fractions having 36 as the denominator in the
following way:
3j_9_27 5^6_30 7 4 _ 28
4 9 ~ 36 ' 6 6 ~~ 36 ' 9 - 4 ~ 36 '
Step 3. Combine the new fractions to obtain
3_57_27_3028 __ 27 - 30 + 28 25
4 6 + 9 ~ 36 36 + 36 ~ 36 ~~ 36 "
Note that adding (or subtracting) the numerators and the denominators of the
given fractions leads to absurdities. Thus, it is not true that ~ + 5 = = Also,
Zoo
dropping the common denominator leads to absurdities. Thus, it is not true thit
3x 2 5x 2
Example 1-36. Express -7-75 -5 --- 1 - - A as a single simplified fraction.
X ~p u A " X X - 4
2 2
Solution: Step 1. Since x 2 - 4 = (x + 2) (x - 2) and - ^ - = - 5 > an
L X X
38 Introductory Topics Sec. 1-26
L.C.M. of the denominators is x 2 4. The given expression may be written
as follows:
3x 2 5x - 2
i O I
x - 2 x 2 - 4
Step 2. When the L.C.M. is divided by the denominators (x + 2), (x 2), and
(x 2 - 4), the results are
5l_ZJ* 2 * 2 ~ 4 - x | o * 2 "" 4 - 1
T+T - * - A x-2 ~ X+2) x - 4 ~ lf
Multiplying the numerators and the denominators of the given fractions by these
factors, we obtain
3x * (a? - 2) 3a 2 - 6a
(x + 2) (x - 2) ~ z 2 - 4 '
2 (s + 2) __ 2j-_4
(a _ 2) Or + 2) " x 2 - 4 '
5x - 2 _ 5a - 2
a; 2 - 4 ~~ x 2 - 4 -
Step 3. Therefore, the desired result is obtained in the following way:
3x 2 __ 5x -2 = (3x 2 - 6g) + (2x 4- 4) - (5x - 2)
z-h2 + o;--2 a: 2 -4 x 2 -4
__ 3o; 2 - 9x 4- 6 3(x - 1) (x - 2) _ 3(x - 1) t
*
x 2 - 4 ~" a; 2 - 4 z+2
EXERCISE 1-8
Perform each of the indicated operations and express the answer in lowest terms.
i 1 2 ^_ 7 9 1 5 7
L 2 ~3 + 8* ~9 ~9*
Q 3 R ,2 xZa.^_A
^7"" &H "2T' ^'S^G 12
. Q ^13 5 5 17 19 t
5 ' "" 3+ 20""Io' 6 '2~T + T
o
- a:
10 1 2,3 1yl a; - 2 z +2
Id. - - r -- - - - + 7^ - ~r- 14.
a; - 1 x 2 - 1 ^ (x - I) 2 a + 2 2 - JK 4 -
x -2 a; +2 x + 3
' .
6 ^ ^2 + 7^ + 12 ^ a: 2 -f 60; + 8
X V Q Q T^ X t x ~r i/ 1 .j
/* /r i 1 Oi > 2 Q < ?'r Q/*
18. x 4. x + A _ zx ~ 6 . 19. * x ^ ,
x + 1^ x x* + x x 2 -y 2 x* -2xy + y 2
Sec. 1-27 Introductory Topics 39
20. 3 _ 1 . 21 .l*,-<>
4 - x 2 x - 2 a -b
22. -^-=-+4 + o^- 23. i+?+A.
x 1 2 x x y xy
T II Til T I/ Til
24 _ __ ^ _j_ * . 25 j_ y ^ .
* x - y x - y x 2 - y 2 ' x +y x - y x 2 - y 2
26 So; 2x 4xi/ f
* x 2 - y 2 x 2 - 2xy + y 2 x 3 - x 2 y - xy 2 -f 2/ 3
1-27. MULTIPLICATION AND DIVISION OF FRACTIONS
The product of two fractions is the fraction whose numerator is
the product of the numerators and whose denominator is the
product of the denominators of the given fractions. That is, by
(1-29), if neither 6 nor d is 0,
n 9cn a . c ac .
(JL4\J) T 1 T~l
b d bd
To illustrate,
26 2-6 4 . x- 1 x + 3 x 2 + 2x - 3
- _________________
3 7~~3-7~7 x + 2 x-2~ x 2 - 4
A special case of (1-29) is worth noting. To multiply a fraction
by a number or expression, we multiply the numerator by that
number or expression. Thus,
b __ a b __ a b _ ab
c ~ 1 c ~~ 1 c ~" c
For example, * / i\
2.| = |, and (-i).K = (^lI^.
55 x x
The quotient of two fractions, T/^J has been defined in
bf a
Section 1-2 and evaluated in Section 1-7. Thus, if no factor in a
denominator is zero,
(1-30) J/5 = J^ = *.
b/ d b c be
To illustrate,
5 /3 __ 5 2 _ 5*2 _ 10 3a? 2 /6a 5 __ 3^ j/^_ _ 3x 2 * y 7 _ j/^_
7/ 2 " 7 ' 3 ~ 7 3 ~ 21 ' 2/ 3 / e/ 7 " 2/ 3 ' 6^ 5 ~ 2/ 3 - Go; 5 "" 2x 3 '
In practice, we divide out all factors common to the numerator
and denominator before proceeding with the actual multiplication.
If necessary, the numerator and the denominator of each given
fraction should be factored.
T 2 __ OT O/>2 Q-y _ u
Example 1-37. Multiply a , , . 9 by r-^V and simplify the result
&x* + &r -p o .u y
40 Introductory Topics Sec. 1-27
Solution: The work may be indicated as follows:
a 2 - 2x 2x* - 3x - 9 _ x(x - 2) (x - 3) (2x + 3)
2z 2 + 5x + 3 ' x 2 - 9 ~ (x + 1) (2x + 3) (x + 3) (x - 3)
_ x(x - 2) = x 2 -2x
~ (x -f 1) (a + 3) ~~ a; 2 -j- 4s + 3 '
n i * oo ^- i x 2 - 5# -f- 4 , rr 3 - 4x 2 + a; - 4
Example 1-38. Divide ^j by ^-^
Solution: By (1-30),
x 2 - 5z + 4 z 3 - 4x 2 + a- - 4 __ (x a - 5x + 4) (2x - 1)
x 2 - 3x - 4 * 2^-1 (x 2 - 3z - 4) (x 3 - 4x 2 + x - 4)
__ (x -l)(x - 4) (2s - 1)
(x + 1) (x -4) (a -4) (z 2 +l)
(s - 1) (2s - 1)
(x + 1) (x - 4) (x + 1)
1-28. COMPLEX FRACTIONS
The fractions we have been discussing so far may be called
simple fractions, to distinguish them from the fractions which we
now discuss.
A fraction which contains other fractions in the numerator or
in the denominator is called a complex fraction. Since the simpli-
fication of a complex fraction is essentially a problem in division,
we first reduce the numerator and the denominator of the complex
fraction to simple fractions and then proceed as in division.
1+ ;
Example 1-39. Simplify- - r
x y
First Solution: The numerator of the given complex fraction reduces to the simple
T I 77 I! \- 7?
fraction - - > and the denominator reduces to - - Hence, we have
x , xy
x +y
(x + y) xy
y -f x x (y + x)
Alternate Solution: Frequently it may be more convenient to multiply both the
numerator and the denominator of the complex fraction by an L.C.M. of the
denominators of all simple fractions occurring in the given complex fraction. In this
example, the simple fractions are - > - > and - > and an L.C.M. of their denomi-
nators is xy.
Therefore, by multiplying the numerator and the denominator of the complex
fraction by xy, we get
\ l + x)' xy __ xy +y* _ (x + y)y _
Sec. 1-28 Introductory Topics 41
EXERCISE 1-9
In each of the problems from 1 to 20, perform the indicated operations and
express the answer as a simple fraction in lowest terms.
1- 5 ' 155 '
* 3 . * .
7 6V
5 14 / 28 .
.2.5.
37
9x 16^ 3
3a; / 16# 3 2
g -U, W , - *,* ~ ^
5 ' 9/45
6a 2 6 3 3a 3 fe
17y 3 / 68y*w
15a 4 6 4 /
Safe 2
54a6 2
8 ' 72r?/ 3 / 16x
,1 9z 4 ?y 3
14a 3 6 2
x 2 - fo + 5
4?/ 27x 3 15z 3
& _ 8 _ ,, + 2x - 3 .
"' x* - 25
30 i />
X 2 4" O
fl 2 2 1 O J_ /I
14 x 4 4- 27z . a; 3 4- 3# 2 - 2x 6
U ' (x-7]
15.
16.
17.
18.
19.
20.
In each of
237
01 3*4*2
i 2 'or 2 5ar 14
27x 3 - 8 4z 2 - 25
x 4 -4.r 2 *(*'-5*+6)
6z 2 + 19x 4- 10 9# 2 - 12;r
6x 2 "1" 5x 21 6x 4 4" 36ic 3
4-4
0^.2 P\T 1 O
^C OX i&
4^.2 9^; 28 9 4# 2
2x 2 -f 5x -h 3 3x 2 - 20x 4
Qr 3 4- 91 r2
t/.*/ 7^ ^>' J */
- 12 6z 2 -f 5a: 6
5^2 24a; 5 ,T 2 4" 3^r 4~
i*2 ^O'y Q^2 r4. / y2 ^'
(/ ~~ \^iU ~~ Oiv^ 1 T:>*/ "~~ ^t
2* J.T2 _L QT _L 9
T:.*- "T" vJs ~\ &
jZ ~~* x) \jv ~~ ^c "~~ &y / i
(T ^^^^ - 4?y^ * L ^2?/ - 7*
T8x 2 - 2x - 15 9 - 4zn
)2 _ g 2 2 (32 _ 2?/) 2 a; 2 J
4x 2 4- 12a: 4- 9
(r 4 - 625) (a: 2 - 9) . 3* 4
48.r 2 7/ 3 4- 60?/ 3
4- 75x 2
(x + 5) 2 (x - 5) 3 ' & + lOx + 25
the problems from 21 to 38, simplify the complex fraction.
5 3 18 123
oo 9'lO'l9. oo 2 3 + 4
" L 5
16
7 4
"" 64
81
3 6xy
""I +1+1
5 ^6 ^8
16a; 2 ^ 3
e 9x
o 4 18 15
10 -5 5 ,
~ * 1 3
1 - 5 4^ 2
" e ' 18^ 3 2/
2 5
f 6 2y
x+--3
-9 *
4x?/
+T- 7
n o a
" ' 2x - y
1 1 4. 7
" J< a 71
y*
6 + 6a;
?-? + '
^1 *
3 3a + 3
1 2
3 ^+2-
' 2 3
4 J 4
of 20
# 8 #
" * 2
' *t - 4
42 Introductory Topics Sec. 1-28
8 x _ 2 1_ g 1 - 2s
33. ^4^- 34. ~2. 35. "" L1
x 1-2*
____ . _
; *v ~
-f 4 # - 4 2# -f 1 x
07
- x 1 +g s +y __ g - y x -2y I", __ 2y
x -lx + l x* -y* x 2 +
Simplify each of the following expressions:
- _ a?
1-29. LINEAR EQUATIONS
Introduction. In Section 1-13, an equation was defined as a state-
ment of equality between two algebraic expressions. In this section
we shall discuss equations in one unknown of the simplest type,
called linear equations, typified by the form
(1-53) , ax = &,
where a and & are specified numbers and a = 0. Some ideas pertain-
ing to equations in general are needed first.
Solution or Root of an Equation. We shall use the term unknown
to designate a literal quantity that appears in an equation and is
not regarded as specified at the outset. A system of values of the
unknowns which, when substituted for them, makes the equation a
true assertion is called a solution of the equation. A solution is
also called a root when only one unknown is involved. Thus, the
values x = 2 and y = 1 define a solution of the equation
3# + 2y = 4, since the equation is satisfied for these values ; that is,
3(2) + 2(-l) = 4.
Sec. 1-30 Introductory Topics 43
Equivalent Equations. Two equations are equivalent if they have
exactly the same solutions. For example, 2x + 11 = 7x - 4 and
4# + 22 = 14# - 8 are equivalent, since, as will be seen, each has
exactly one root, namely, x = 3. However, 2x + 11 = Ix - 4 and
2# 2 + llz = To; 2 - 4# are not equivalent, because the latter equation
has, in addition to the root x = 3, the root x = 0.
Operations on Equations. Each of the following operations on an
equation yields an equivalent equation :
Adding the same expression to (or subtracting the same expres-
sion from) both sides.
Multiplying (or dividing) both sides by the same non-zero
number.
For any solution of F = G also makes F + H = G + H true, and
is therefore a solution of this latter equation. Conversely, any
solution of F + H = G + H is one of F = G, because
F = F + H-H = G + H-H = G.
Likewise, any solution of Fj= G is one of aF = aG, provided that
a is a non-zero number, and conversely.
Transposition of Terms. Transposing a term of an equation con-
sists in moving the term from one side of the equation to the other
and changing its sign. This operation is equivalent to adding the
same quantity to both sides (or subtracting the same quantity from
both sides) . For example, consider the equation
2x+ 11 = 7x - 4.
If we transpose 2x from the left side to the right, and transpose
-4 from the right to the left, we obtain
11 + 4 = Ix - 2x.
In effect, we subtracted 2x from each side and added 4 to each side.
1-30. LINEAR EQUATIONS IN ONE UNKNOWN
An equation in the form shown in (1-53) is called a linear equa-
tion in one unknown. We shall show that such a linear equation
has one and only one root, namely,
(1-54) x = -
a
If each side of the equation ax = b is divided by a, the result is
b
x = - >
a
44 Introductory Topics Sec. 1-30
which is equivalent to the given equation. Therefore, (1-53)
clearly has one and only one solution, namely, b/a.
It should be noted that if we allow a to be in (1-53), there are
two possibilities. Either no solution exists, when 6^0, since for
no number x is it true that (fx = Q 9 x = Q = b^Q; or else every
number # is a solution, when 6 = 0, since a*# = 0'# = = & for
all values of x.
An equation which is not apparently linear may frequently be
solved by the theory of linear equations, by replacing the given
equation by a linear equation to which it is equivalent. The follow-
ing steps serve as a guide to the method to be used when there is
only one unknown in the equation.
1. Clear the equation of any fractions with numerical denomina-
tors by multiplying both sides of the equation by a least common
multiple of the denominators of those fractions.
2. Transpose all terms containing the unknown to one side and
all other terms to the other side. We may collect the terms contain-
ing the unknown on either side.
3. Combine like terms. If the equation now assumes the form
in (1-53), it is a linear equation and can be solved by dividing both
sides by the coefficient of the unknown.
4. Check the result obtained in step 3 by substituting in the
original equation. While it is desirable to include step 4 to show
that the number x found in step 3 is actually a solution of the
equation, step 4 is not a necessary part of the solution process,
since the operations performed in the preceding steps always
yield equivalent equations. The purpose of step 4 is to help
to make certain that there has been no error.
In step 2, we generally transpose terms containing the unknown
to whichever side makes solving the equation easier.
The following explanations should be noted carefully.
Students at times "transpose" coefficients of the unknown. Thus,
2x = 6 takes the erroneous form x = 6 2 by "transposing" 2 to
the right side. The correct procedure is to remove the coefficient
2 by division, since it is a multiplier of x. Therefore, we should
divide both sides of the equation 2x = 6 by 2 and have
2x 6
or
2 = 2'
a; = 3.
Errors of this type may be avoided if the student applies the rules
of algebra properly and checks his solutions carefully.
Sec. 1-30 Introductory Topics 45
Multiplying or dividing both sides of an equation by a poly-
nomial involving the unknown will not necessarily yield an equiva-
lent equation. When the operation is multiplication, the new equa-
tion thus obtained may have roots in addition to the roots of the
original equation. These extra roots are called extraneous roots.
We then say that the equation is redundant with respect to the
original equation. When both sides of an equation are divided by
a polynomial involving the unknown, the new equation may lack
some of the original roots. It is then said to be defective with
respect to the original equation.
If both sides of an equation are multiplied by a polynomial
involving the unknown, the check in step 4 of the recommended
procedure is a necessary step in the solution process. An extra-
neous root can thus be identified. Dividing both sides of an equa-
tion by a polynomial involving the unknown is not a permissible
procedure, since roots that are lost cannot be regained.
It should be noted that a non-linear equation may some-
times be treated so as to obtain a linear equation which is possibly
redundant. For example, the fractions in the equations of Problems
39 to 48 of Exercise 1-10 may be eliminated by multiplying both
sides of each equation by a least common multiple of the denomina-
tors of the fractions in that equation. Since this multiplier con-
tains the unknown, the new equation may have solutions which are
not solutions of the given equation. Hence, the solutions must be
checked to see whether or not they actually are solutions of the
given equation.
ST x A 7
Example 1-40. Solve the equation ~- 4 '=
o o
Solution: We clear the equation of fractions by multiplying both sides by 15,
which is an L.C.M. of the denominators of the fractions, and obtain
Wx = 60 - 3x - 21.
Collecting the terms containing x on the left side and all other terms on the
right, we have 10* + 3* = 60 - 21.
Combining like terms, we obtain
I3x = 39.
Finally, dividing both sides by 13, we have
| * = 3.
To check, substitute 3 for x in the original equation. The result is
?l2 - 4 - iZ or 2=4-2, or 2=2.
o o
Therefore, 3 is the root.
46 Introductory Topics Sec. 1-30
Example 1-41. Solve the equation 6x - 3y - 1 = 5y -f- 2x -f 11 for y in terms
of x. In this case, regard x as specified.
Solution: Collect the terms in the unknown y } on the right side, and collect all
other terms on the left. The result is
6x - 2x - 1 - 11 = by + 3y.
Combining like terms, we have
4z - 12 = %.
Now we divide both sides by the coefficient of the unknown to obtain
$x - 2 _
g -IT
Then ' --
# 3
To check, substitute for ?/ in the original equation, and obtain
^ At
This equation reduces to
I2x - 3(x - 3) - 2 = 5 (a? - 3) + 4z + 22,
which becomes
I2x - 3z + 9 - 2 = 5z - 15 + 4^ + 22,
or
9x + 7 = 9x + 7.
Since this result is an identity, y = ^ is the solution of the original
equation, regardless of the value of x.
Example 1-42. Find two consecutive integers such that four times the first is
equal to six times the second diminished by 20.
Solution: Let x be the smaller integer, and x + 1 the next larger integer. Then,
from the statement of the problem, we have
4z = 6(x 4- 1) - 20.
From this equation, we obtain
4z = 6z -f 6 - 20.
Then,
14 = 2x, or s=7.
Hence x = 7 and x -f 1 = 8 are the two consecutive integers. The student
should carefully check these values by substitution in the original statement of
the problem.
Example 1-43. The speed of an airplane in still air is 400 miles per hour. If it
requires 20 minutes longer to fly from A to B against a wind of 50 miles per hour
than it does to fly from B to A with the wind, what is the distance from A to ?
Sec. 1-30 Introductory Topics 47
Solution: Let x be the distance from A to B. From the data, the speed of the
plane against the wind, in miles per hour, is 400 50 = 350, and the speed of the
plane with the wind, in miles per hour, is 400 4- 50 = 450.
Since distance = rate time, or d = rt, we have - = t. Hence,,
;rr
OOU
and
;rrrr = time, in hours, required to fly from A to B,
jrr = time, in hours, required to fly from B to A.
Therefore, from the statement of the problem, it follows that
x x _ 1
350 ~450 "~3*
Solving this equation, we have
9z - 7x = 1050, or x = 525.
So the distance from A to B is 525 miles.
EXERCISE 1-10
Solve for the unknown in each problem from 1 to 15.
1. - 4z - 2 = 3 - 2x. 2. 3x - 6 = x + 12. 3. 3x - 2 = - 4z - 5.
C r K 1
4. 30 + 7 = 2 - 20. 5. - 9x - 7 = 6 - 6. 4w + = 3w - i
J L i
7. 4 - 3z = 6(1 + 2). 8. 4z - 6 = 3z 4- 4. 9. x - 8 = 2x + 3.
10. 60 + 7 = 50 + 6. 11. lOx - 3 = 9z + 4. 12. \x - | = | - jx
4 ju ^ 4
13. 5z - 7 - 8z = 4x - 17 - fcc.
14. Hty + 4 + 6y = 2y + 7 + 3y. 15. 6(5 + 4*) - 3(x - 4) = 0.
Solve for y in terms of x in each problem from 16 to 24.
16. 2x - y = 3. 17. 23z + 20 = 4. 18. a: - 40 + 10 = 0.
19. 40 - 2x + 2 = 0. 20. - + 1 = 1. 21. 6z - 20 = 3.
Q> *
22. 3(x - 4) + 40 = - 3. 23. 2x + 4(0 - 3) = 5. 24. 3x + 7(0 - ij = | .
Solve for all values of x which satisfy the equation in each problem from 25 to 48.
25. 5z - 3 = 4(s - 2). 26. 8fo - 2^ - 9( - 4) = 13. 27. !^-J> = n.
o s 23 + 17
n 9 - 4x 4 OA 5 + 6z
40
.8. 3 7.
3+5 x + 7 _ q
"A Q ' Q
O o o
^-6 3-8 33 3+3 J
23-7
31- 4 5 3
, 23 - 1 a; - 5
.2^+3 33-1 73 + 3
12
2 x I !
34. 3 7 < J
,_ 83 - 21 1 53
4 2 6
OQ 7 - 3 43+3 63
5 a;+ 2
16
J7 ' 5 4 ~ 6
39. 3 4 - 6^ 4> \
J3 ' 4 7 5
40. . 4 _ 4-3-7.
3
48 Introductory Topics Sec. 1-30
41
4f
u
42,
X
X
A X
X
#2 %
S 2
S
9
9 s
43.
4 ,
5
1
44.
1 +
3
4
2x
+ 3 + s
-
4 4^E 2 ..,
10s
-24
s -
1 s
+ 1
45.
3
1
46.
-
1
19 -
22s
6s
2 -2s +
1
2s 2 - 4s
+ 7
x +
2
s ~* s
- 6
47
5
-i
7 -s
5 -2s
-,T 2
- 3
- rr 2
2(x 2 + 3rc + 9) 81 - ~~
49. Divide 98 into two parts such that one of them exceeds the other by 18.
50. Find three consecutive integers whose sum is 84.
51. Find two consecutive integers whose squares differ by 13.
52. If 8 times a certain number is 9 more than 5 times the same number, what is
the number?
53. A rectangular plot of ground is four times as long as it is wide. If its perimeter
is 4,800 feet, what is its area?
54. How many pounds of coffee at 90 cents per pound and how many pounds at
98 cents per pound will it take to make 100 pounds of a mixture costing 96
cents per pound?
55. At a college play, admission was 25 cents for a child and 75 cents for an adult.
If $210 was taken in from 500 admissions, how many children and how many
adults were admitted?
56. A man has $365 in 41 bills of $5 and $10 denominations. How many bills of
each denomination does he have?
57. What are the angles of a triangle, if one angle is three times the second angle
and six times the third angle?
58. One man, X y can do a certain job in 7 days, and another man, Y, can do the
same job in 15 days. How long would it take them to do the job working
together?
2
The Function Concept
2-1. RECTANGULAR COORDINATE SYSTEMS IN A PLANE
In Section 1-5 we saw how we can associate a real number with
every point on a number scale. The real number attached to a given
point is called the coordinate of the point. This representation sug-
gests the assumption that to any real number there corresponds
precisely one point on the scale, and to any point of the scale there
corresponds precisely one real number. This one-to-one correspond-
ence between the set of real numbers and points on the number
scale is known as a one-dimensional coordinate system.
We shall now extend the concept of a one-dimensional coordinate
system to a system of coordinates in a plane in which two number
scales are perpendicular to each other. The two perpendicular
lines, which we shall call coordinate axes, divide the plane into four
parts, or quadrants^ numbered as shown in Fig. 2-1. The horizontal
and vertical lines are designated as the
x-axis and the y-axis, respectively, and
their point of intersection is called the
origin and is labeled O. D
On each of these axes, we construct a
number scale by selecting an arbitrary
unit of length and the origin as the zero
point. As in Section 1-5, a coordinate on
IV
the x-axis will be considered positive if
it is to the right of 0, that is, to the
right of the #-axis, and will be negative p IGt 2-1.
if it is to the left. A coordinate on the
2/-axis will be considered positive if it is above the #-axis, and nega-
tive if it is below.
Just as the real-number scale of Fig. 1-1 gave us a system of
one-dimensional coordinates by which we could set up a one-to-one
correspondence between points on a line and real numbers, so the
49
50 The Function Concept Sec. 2-1
system of coordinates with respect to two mutually perpendicular
axes sets up a one-to-one correspondence between points in a plane
and ordered number pairs. We use the designation "ordered" pairs
for the following reason. To designate any point, we shall agree to
give its directed distance frdm the i/-axis first, and call it the
abscissa or x-coordinate, and then the directed distance from the
#-axis and call it the ordinate or y-coordinate. The abscissa and
ordinate of a point constitute its rectangular coordinates. They are
written in parentheses as an ordered number pair, as in the nota-
tion (x, y), the abscissa always being written first. By this scheme
we assign to each point of the plane a definite ordered pair (x, y)
of real numbers and, conversely, to each ordered pair (x, y) of
real numbers there is assigned a definite point of the plane.
Thus, the abscissa of the point P in
Fig. 2-2 is 3, and its ordinate is 2, and
we say that the coordinates of the point
PO.2) P are (3, 2). Similarly, the coordinates
T of Q are (2, 0), the coordinates of R
\ \ \ +-x are (0, 3), and those of the origin are
(0, 0). Marking in the plane the posi-
tion of a point designated by its coordi-
nates is called plotting the point.
The coordinate system we have con-
FlG 2-2 stf ucted is a particular case of cartesian
coordinates, so called in honor of Rene
Descartes (1596-1650), who first introduced a coordinate system
in 1637. It is called a rectangular system, since the axes intersect
in a right angle. (Actually the axes may intersect at any angle, but
it is usually simpler to take them perpendicular to each other. When
the two axes are not perpendicular, the coordinate system is called
an oblique coordinate system. Oblique systems will not be used in
this book.)
2-2. DISTANCE BETWEEN TWO POINTS
It was seen in Section 1-10 that |a 6| equals the distance
between two points on the number scale represented by the real
numbers a and 6. It follows that \x 2 - a?i| represents the distance
between the points A (xi, 0) and B(x 2 , 6) on the #-axis of Fig. 2-3.
Let us now consider the two points PI and P 2 with coordinates
(#i> #1) and (# 2 , yi). The points have the same ^/-coordinate, which
means that they lie on the same horizontal line. Hence, the dis-
tance between the points PI(XI, yi) and P 2 (tf2> 2/i) is the same as
Sec. 2-2
The Function Concept
51
*K
*-JT
FIG. 2-3.
FIG. 2-4.
the distance between A(#i, 0) and Z?(# 2 ,0). This distance is
| #2 ~ #i|- Similarly, we can show that [2/2 2/i| represents the dis-
tance between two points (# 2 , 2/1) and (# 2 , 2/2) on a vertical line.
We have thus arrived at the following two important properties.
If two points PI (#1,1/1) and P 2 (# 2 , 2/i) have the same ^-coordi-
nate, then the distance between them, or |PiP 2 |, is given by
(2-1) |PiP 2 | = |*2 - xi\.
If two points Qi (#1,2/1) and Q 2 (#1,2/2) have the same #-coordi-
nate then the distance between them is given by
(2-2) IQiQal - (2/2 ~ l/i|.
The concept of the distance between any two points in a plane is
so important that we shall now develop a formula for it. Let us
denote by d the distance between the points PI (#1,2/1) and
P 2 (#2, 2/2). That is, d is the length of the line segment PiP 2 in Fig.
2-4. Let P 3 be the point (# 2 , yi) as shown.
Since the angle at P 3 is a right angle, we have, by the Pytha-
gorean theorem,
|PlP 2 | 2 - |PlP 3 | 2 + |P 3 P2| 2 .
Therefore,
|PiP 2 | 2 = |*2 - #i| 2 + It/2 - yi\ 2
= (#2 - #i) 2 + (2/2 ~ 2/i) 2 .
and
That is, the distance d between any two points PI (#1,2/1)
P2(#2, 2/2) in the plane is given by
(2-3) d = V(#2 - #i) 2 + (2/2 - 2/i) 2 -
This formula is known as the distance formula.
Example 2-1. Find the distance between the points ( 3 2) and (5, 2).
Solution: It makes no difference which point is labeled Pi. Let us label the first
one Pi and the second one P 2 . Since the two points have the same ^-coordinate,
(2-1) applies and |PiP 3 1 = |5 - ( - 3)| = |5 + 3| = |8| = 8.
52 The Function Concept Sec. 2-2
Example 2-2. Find the distance between the points (3, 1) and (3, 7).
Solution: Again let us label the first point PI and the second one P 2 . Since the
points have the same ^-coordinate, (2-2) applies and |PiP2| = |7 1| = 6.
Example 2-3. Find the distance between the points (2, - 1) and (5, 3).
Solution: Let us designate the points as Pi(2, - 1) and /Vo, 3). By (2-3),
the distance |PiP2| is
-2)* +(3 -(-I)) 2
= V9 + 16 = \/25 = 5.
We shall now consider a special application of the distance
formula (2-3). The line segment OP from the origin to a point
P(x,y) is called the radius vector to P. By (2-3), the distance
between O(0, 0) and any point P(x,y), or the length of the radius
vector to P, is
or d = V(* ~ O) 2 + (y - O) 2
(2-4) d = V* 2 + y 2 -
Thus, for the point P(3, 2), the radius vector has the length
V3 2 + 2 2 = VIST
EXERCISE 2-1
1. Plot the following points:
a. (3,5). b. (-4,7). c. (5, -2).
d. (-3, -6). e. (1, -3). f. (-8, -6)
g. (0,2). h. (-5,0). i. (3,0).
2. In each of the following cases, plot the pair of points and find the distance
between them:
a. (2, 3) and (7, 3). b. (5, - 2) and ( - 1, - 2).
c. (1, 4) and (1, 0). d. ( - 3, - 2) and ( - 3, 4).
e. (2, 1) and (5, 6). f. (0, - 8) and (5, - 3).
g. (3/2) and (5, 7). h. (- 4, - 6) and (- 8, - 6).
3. Find the length of the radius vector to each of the following points :
a. (4, 3). b. (12, 5). c. (1, - 1).
d. (5, -12). e. (7,0). f. (-3, -2).
g- (0, 4)._ h. (-1,2). i. (a, 6).
j- (1, V3). k. (- \/, - 1). L (m, n).
2-3. FUNCTIONS
So far we have been concerned with single numbers and pairs of
numbers. Now we shall consider mathematical relations, known
Sec. 2*3 The Function Concept 53
as functions, between two sets of numbers. To distinguish between
the two sets, we shall call one of these sets the domain of definition
X of the function, and the other the range set Y of the function.
We begin by defining a variable to be a symbol which may take
any value in a given set of numbers. If # is a symbol which is used
to denote any number of the domain X and y is a symbol which
denotes any number of the set Y, then x is called an independent
variable and y is called a dependent variable. If a set contains only a
single number, the symbol used to represent that number is called
a constant.
To set forth a function, the domain should be explicitly specified ;
that is, it is necessary to determine definitely just what elements
or numbers the domain contains. The same is true of the range set
Then, as soon as a definite rule of correspondence is given which
assigns to each number x of the domain one or more numbers y of
the range set, the function is specified completely. We thus have a
set of ordered pairs of numbers (x, y), where x is any number of X
and y is a number of Y.
The set of ordered pairs of numbers (x,y) is called the function.
The rule of correspondence which determines the collection of pairs
(x,y) is often expressed by a formula involving algebraic or other
processes. In .such cases we usually find it convenient to refer to the
formula as though it were the function. For example, we often
speak of the function y = x 2 3x + 5 when actually we mean the
set of ordered pairs (x,y) determined by y = x 2 3x + 5. If just
one number of Y is paired with each value of x, the function is said
to be single-valued. If more than one number of Y is paired with
some value of x, the function is said to be multiple-valued. We fre-
quently find it possible to deal with multiple-valued functions by
separating them into distinct single-valued functions.
The range of values of the function consists of those numbers y
in the range set Y which actually correspond to some number x of
the domain. When the range of a function has been determined,
it is always possible to replace the original range set, which may
include numbers in addition to those of the range, by the range
itself. We shall now consider several examples of functions.
Illustration 1. The constant function y c associates the same
number c with every number x of the domain X. Hence, the range
Y of the function consists of just one number c. Since y has the
same value for all pairs (x, y), the function is evidently single-
valued.
Illustration 2. The identity function y = x associates with every
real number x the number itself. In other words, the numbers
54 The Function Concept Sec. 2-3
corresponding to x = 1, 2, 3, are y = 1, 2, 3, respectively. The
domain X is the set of all real numbers, and the range Y is also
this entire set. The function y = x is single-valued, because it asso-
ciates just one number of Y with each value of x.
Illustration 3. Let us consider the linear function defined by the
equation y = 3x 2. The domain X is the set of all real numbers,
and the range Y is also this entire set. In this case a given x deter-
mines a unique y which is equal to 3x 2. For example, corre-
sponding to x = 1, 2, 3, we have y = 1, 4, 7. The function is single-
valued.
Illustration 4. Let the rule of correspondence be given by the
equation y = x 2 . Also, let X be the set of all real numbers, and let Y
be the set of all non-negative real numbers. Then, to the number
x 2 there corresponds the number y = ( 2) 2 4; to the number
x = 3 there corresponds the number y = 9 ; and so on. Hence, y = x 2
associates just one number of Y with a given value of x, and defines
y as a single-valued function of x. Although this is not obvious, the
range of the function is the given range set Y.
Illustration 5. In this case let the rule of correspondence be given
by the equation y 2 = x. Here X is the set of all non-negative real
numbers, and Y is the set of all real numbers. Then to the number
x - 2 there correspond the two numbers y \/2 and y = \/2; to
the number x = 9 there correspond the numbers y = V9 3 and
y = - V9 = ~3 ; and so on. Thus, y 2 = x defines y as a two-valued
function of x. Only for x = is there a single value of y, namely,
y = 0.
Illustration 6. The function y = \A& 2 # 2 > with domain X con-
sisting of all real numbers x such that a ^ x ^ a, is a single-
valued function with range set Y consisting of the numbers
^ y ^ a. Here the range is Y, as may be proved.
Note that \fa 2 x 2 has no meaning when a 2 x 2 is negative.
Hence, those values of x must be excluded for which a 2 x 2 < 0,
and we must restrict the value of x to the interval a ^ x ^ a.
(We shall consider the meaning of the square root of a negative
number in Chapter 11.) The function y = -\/a 2 x 2 is single- valued
because \/a 2 x 2 represents the non-negative square root of a 2 x 2 .
(The other square root of a 2 x 2 is denoted by V a2 ~~ # 2 -)
Illustration 7. The function y = l/x is defined for all real num-
bers different from 0. For x = 0, l/x is not defined, since division
by zero is not permissible. In other words, although there are
Sec. 2-4 The Function Concept 55
values of y for values of x neat* 0, there is no possible value for y
when x actually equals 0.
Usually the functions which we are about to consider are defined
for all values of x 9 with the following two exceptions :
Values of x must be excluded which involve even roots of
negative numbers, since these are not defined as real numbers.
Values of x must be excluded for which a denominator is zero,
since division by zero is not a permissible operation.
On occasion, the use to be made of a function will restrict the
values of x for which it is to be regarded as defined.
2-4. FUNCTIONAL NOTATION
Since functions are mathematical entities, they may be given
letter notations, such as /, #, <>. To designate the number, or num-
bers, y corresponding to a given number x according to the rule
specified by a given function /, we us6 the notation f(x). As an
illustration, let the function / be defined by the equation
y =* x 2 - 2x + 3.
Then /(O) = 3, /(-I) = 6, and so on. Frequently, the symbol f(x)
is used to designate the function rather than the functional values.
The context will make the meaning cleai*.
It should be remembered that the notation y = f(x) does not
mean that y is a number / multiplied by another number x. Instead
it is an abbreviation for "/ of x."
The set X does not have to be as simple as in the preceding illus-
trations. If X should consist of a set of ordered pairs of numbers, the
rule of correspondence would then determine a value, or values, of y
for each ordered pair of X. We would then have a function of two
independent variables. For example, the area of a triangle is given
by the relationship
A = /(a,6)=|a&.
Here X is the set of ordered pairs, (a, 6), of positive real numbers,
where a is the length of the altitude of the triangle and 6 is the
length of its base ; and Y is the set of numbers A, each of which
represents an area corresponding to a given pair (a, 6). Similarly,
a function of three variables, f(x,y,z), is defined in terms of a set
X of ordered triples (x,y,z). Thus, if f(x, y) = x 2 + y* 9 then
/(2, 3) = 2 2 + 3 2 = 13. Also, if f(x, y> z) ~ x - y + 2*, then / (3, 2, 5)
= 3-2 + 2(5) =11.
56 T/ra Function Concept Sec. 2-4
EXERCISE 2-2
By using the phrase "a function of" in each problem from 1 to 8, express each
given quantity, which is regarded as a dependent variable, as a function of one or
more independent variables. Where possible write the relationship both in words
and in symbols.
1. The area of a circle. 2. The area of a triangle.
3. The area of a trapezoid. 4. The volume of a sphere.
5. The volume of a cylinder. 6. The retail price of food in a grocery store.
7. The annual premium for a life insurance policy.
8. A person's height.
9. Given f(x) = 2x - 3, find /(O), /(- 1), /(3), /(1/2), /(V), 3/(l), /(3)/4,
10. Given g(x) = 3x + 5, find 0(1), g(- 3), (0(2)), 0(4), 20(4), g(z - 1).
/( 2)
11. Using f(x) and 0(x) as defined in problems 9 and 10, find j^- > /(6)0(3),
In problems 12 to 26, let /(a;, j/) = 2xy + 3.t - 2y, and let 0(a, 6, c)
= a 2 + 6 2 + c 2 . Evaluate or simplify each given expression.
12. /(I, 2). 13. /(O, 0). 14. 0(0, 0, 0).
15. 0(1, 2, 3). 16. /(i, 1/x). 17. p(p, g, r).
18. M if) + ffe V, -). 19- 20.
25. g(- a, 6, c) - g(a, - 6, c). 26. 0(a, 6, - c).
27. Express the area A and circumference of C of a circle as functions of the radius r.
By eliminating r, express A as a function of C, and also express C as a func-
tion of A.
28. Express the volume of a sphere as a function of its surface area.
29. Express the surface arfea of a sphere as a function of its volume.
30. Suppose that U is a function of V and that V is a function of W. Show that U
is a function of W.
Determine the maximal domain of values of x for which y is defined as a function
of x in each problem from 31 to 61. Assume that x and y are real numbers.
31. y = x. 32. y = 3z. 33. y = - |
34. y == 3z + 1. 35. y = 2x - 3. 36. y = 4z + 5.
37. y = z 2 . 38. y = x 2 - 2. 39. y = x 2 + 1.
40. y = #(2z -f- 1). 41. y = (3x- 1) (a? 4- 1). 42. y = (2x - 3) (2x + 1).
43. y=x(x-l)(x+l). 44. 2/=z2+i- 45. y = l
3/ X\X L)
46. x 2 + y 2 = 9. 47. x 2 - 2/ 2 = 9. 48. x 2 - 2/ 2 = 0.
49. x 2 + = 0. 50. 3* 2 + tf = 0. " * 2
Sec. 2-6 The Function Concept 57
54.,=
55 - y = -T7 ' 56. z = 4i/2. 57. s + 1 = Zy*.
x T- i
58. y = Vr^2. 59. y = V4 - a: 2 . 60. y =
2-5. SOME SPECIAL FUNCTIONS
The following additional illustrations of two rather unusual, but
very useful, functions are given here to help us become better
acquainted with the function concept.
The absolute-value function is defined by associating with x its
absolute value x\. The functional relation is given by j/=|a?|.
Thus, the domain comprises the set of all real numbers, while the
range comprises the set of all non-negative real numbers. For this
function we have the values 2, 3, 0, TT, \/2 corresponding, respec-
tively, to x = 2, -3, 0, 77, -\/2.
The bracket function or greatest-integer function, represented
by the notation y = [x] , is defined as the largest integer which does
not exceed x. Its domain is the set of all real numbers, and its
range is the set of all integers. Thus, if f(x) = [>], then /(-3.5)
= -4, /(-I) - -1, /(O) - 0, /(2.5) - 2, and /(5) = 5.
EXERCISE 2-3
1. Given /W = fa? | , find/(- 3),/(2.3), an
2. Given /(x) = [x], find/(3.2),/(2), and/(- 3).
3. Given /(x) = x r [x], find/(- 3) and/(2.5).
4. Given /(x) = | x \ + [x] - x, find/( - 3.5) and/(4).
5. Given f(x) = [x] + [2 - x] - 1, find/(0),/(- l/2),/(l),/(3/2), and/(2).
6. Let/(x) be the function whose domain X is the set of all real numbers for which
the definition is as follows :
if x < 0, then/(aO = - x\
\ix ;>0, then/(x) = x.
Find/(3)and/(-2).
2-6. VARIATION
A particularly important example of a simple type of function
often occurring in the physical sciences is given by the formula
y = kx.
If k > 0, this equation shows that y increases a$ x increases, and
that y decreases as x decreases. We usually say that y varies directly
58 The Function Concept Sec. 2-6
as x, or that y is directly proportional to x. If x ^ 0, the relation-
ship may also be written as follows :
where k is called the constant of proportionality. This relationship
is equivalent to saying that the ratio of y to x remains constant for
all non-zero values of x.
The value of the constant k in any particular problem may be
determined from a known pair of values of x and y in the problem.
Thus, if the given relationship is y = kx, and if we know that y = 6
when x = 2, then k = 3. Th formula then becomes y = 3x.
We say that y varies inversely as x, or y is inversely proportional
tO a, if r.
V = I (**<>>
C
This relationship shows that ?/ decreases as x increases, and that y
increases as x decreases. But, when x = 0, the following two for-
mulas are equivalent : ^
xy = fc and y = -
*c
Therefore, the relationship between a? and # is such that the product
of x and y is constant.
Several types of variations may be combined in a single equation
to express a certain law. For example, when y varies directly as x
and z, we say that y varies jointly with x and z 9 and we write
y = kx&.
Direct variation and inverse variation are often combined in
applications. Thus, according to Newton's law of gravitation, the
force F of attraction between two bodies of masses m\ and w 2 varies
directly as the product of their masses and inversely as the square
of the distance d between them. The equation is
_ __ kmim,2
* ~~d?~'
Example 2-4, If a man is paid $15 for an 8-hour day, how much would he make
in a 35-hour week?
Solution: The wages a man earns vajy directly as the amount of time he works.
Since this is a problem in direct variation, we have
w = kt.
In this formula, w represents the total wages, in dollars; t represents the time
worked, in hours; and the constant k represents the wage rate in dollars per hour.
Substituting w = 15 and t = 8 determine^ the constant k = 1.875. The general
formula then becomes
w = 1.875 t.
Sec. 2-6 The Function Concept 59
Therefore, when t = 35, we have
w = (1.875) (35) = 65.62.
Hence, the man earns $65.62 in a week.
Example 2-5. A motorist traveling at an average rate of 50 miles per hour made
a trip in 5 hours. How long would it take him to make the same trip at an average
rate of 60 miles per hour?
Solution: Since the time required Varies inversely as the speed, we have
'=*
In this case, k represents the distance traveled, in miles; r is the speed, in miles
per hour; and t is the time required, in hours. Substituting t = 5 and r = 50
determines k = 250. Hence, the general formula is
, _250
t
r
Therefore, when r = 60, we obtain
250 _ 25
' ~ 60 ~ 6 '
Hence, the time required is 4 hours 10 minutes.
Example 2-6. Under suitable conditions the electric current / in a conductor
varies directly as the electromotive force E and inversely as the resistance R of the
conductor. When E = 110 volts and R = 10 ohms, / = 11 amperes. Find what
voltage is necessary to cause 2 amperes to flow through 60 ohms of resistance.
Solution: From the statement of the problem, we see that the combined variation
is given by the formula fejjj
1 ~~W'
Substituting I 11, E = 110, and R = 10 determines the constant k to be 1.
Therefore, the formula becomes
/=
R
This relationship may also be expressed as follows: The required voltage E
is equal to the current / flowing through the conductor multiplied by the resistance
R, or E = IR.
For the specified values, E = (2) (60) = 120. Hence, E = 120 volts.
In this problem k = 1. The formula obtained is commonly known as Ohm's law.
It is widely applied to entire and partial circuits through which elefctric currents flow.
EXERCISE 2-4
1. If y varies directly with x, and y = 15 wh&i x = 7, find a formula for y in
terms of x.
2. If x varies directly with y, and x *= 32 when y = 4, find x When y = 3.
3. If y is directly proportional to x 2 , and y = 112 when x = 4, find y when x = 9.
4. If y is inversely proportional to x, and y = - % when x = 1, find y when x = -3.
60 The Function Concept Sec. 2-6
5. If y varies inversely with x, and y = 10 when x = 3, find y when x = 6.
6. If # varies directly with y and inversely with 2, and # = 4 when y = 12 and
2=2, find x when ?/ = 16 and 2=4.
7. If y varies directly with \/x and inversely with 2 2 , and y = 18 when x = 9
and 2=2, find / when z = 25 and 2=6.
8. If y is directly proportional to x and inversely proportional to \/z, and 2/ = 4
when # = 1 and 2 = 1, find y when x = 2 and 2=4.
9. If y varies directly with x 3 and inversely with 1 2 2 , and y = 2 when # = 1
and 2=2, find y when a: = - 1 and 2 = - 2.
10. If two spheres have radii r\ and r 2 , diameters d\ and ^2, and surface areas Si
and 2 , respectively, show that
ri* d Si.
r 2 2 ~" d 2 2 "~ Sa *
11. If the two spheres in problem 10 have volumes V\ and F 2 , respectively, show
that
12. If the radii of two spheres are 3 units and 1 unit, respectively, find a) the ratio
of their surface areas and b) the ratio of their volumes.
13. What are the ratio of the surface areas and the ratio of the volumes of two
spheres if the ratio of their radii is 3/2?
14. The diameter of the planet Jupiter is approximately 10.9 times the diameter of
Earth. Assuming that both planets are spheres, find a) the ratio of their surface
areas and 6) the ratio of their volumes.
15. The diameter of the sun is approximately 109 times the diameter of Earth.
Compare the volumes and the surface areas of the sun and Jupiter, if we assume
that the sun and both planets arc spheres.
16. By how much must the diameter of a sphere be multiplied to give a sphere
whose surface area is 25 times that of a given sphere?
17. When the volume V of a gas remains constant, the pressure P varies directly as
its absolute temperature T. (Absolute temperature is measured from the
so-called absolute zero, which is approximately - 460 F or - 273 C.) If gas
is enclosed in a tank having a volume of 1,000 cubic feet and the pressure is_
54 pounds per square inch at a temperature of 27 C, what will be the tempera-
ture when the pressure is raised to 108 pounds per square inch?
18. If the temperature T of a gas remains constant, the pressure P varies inversely
as the volume V. A gas at a pressure of 50 pounds per square inch has a volume
of 1,000 cubic feet. If the pressure is increased to 150 pounds per square inch
while the temperature remains constant, what is the volume?
19* The weight of a body above the earth's surface varies inversely as the Square
of the distance from the center of the earth. If a certain body weighs 100
pounds when it is 4,000 miles from the center of the earth, hdw much will it
weigh when it is 4,010 miles from the center arid when 4,100 miles from the
center? .
Sec. 2-7 The Function Concept 61
20. The electrical resistance of a wire varies directly as its length and inversely as
the square of its diameter. A copper wire 10 inches long and 0.04 inches in
diameter has a resistance of 0.0656 ohms, approximately. What is the resistance
of a copper wire 1 inch long and 0.01 inches in diameter?
21. What is the diameter of a copper wire 1,000 inches long whose resistance
is 10 ohms?
22. According to Kepler's third law, the square of the time it takes a planet to
make one circuit about the sun varies as the cube of its mean distance from the
sun. The mean distance of the earth is 92.9 million miles, and the mean
distance of Jupiter is 475.5 million miles. Find the time it takes Jupiter to
make one circuit about the sun.
j
2-7. CLASSIFICATION OF FUNCTIONS
It is often desirable to group functions into classes. For our
immediate purpose it will suffice to consider a classification into
algebraic functions and non-algebraic, or transcendental, functions.
Let us first give a more precise definition of a polynomial function
and then define algebraic functions and give some illustrations
of both.
A polynomial function of # is a function given by the relationship
y = a G x n + aix n ~ l + + a n _iz + a n ,
where a , a lf - , a n -i, a n are real constants, a ^ 0, and n is a posi-
tive integer or zero. The polynomial function is said to be of
degree n. The function which makes the number correspond to
every number x is also called the zero polynomial, but this poly-
nomial has no degree.
A rational function of x is a function which either is a poly-
nomial function or can be expressed as a quotient of two poly-
nomials. Thus, a polynomial is often referred to as a rational
integral function of x.
A polynomial, or a rational integral function, of x, y, z, , is
defined to be the algebraic sum of terms of the form
kx a y b z c - ,
where k is a constant coefficient and each of the exponents a, b, c,
- - - is either a positive integer or zero*. The degree of such a func-
tion is the degree of the highest-degree term which is present.
For example, the expressions 3x 2 - 5 and 5x 2 - Ixy* + 62 define
polynomial functions of the second degree and third degree, respec-
tively. These and the expressions x 7 y and 2x - \/7 + 9 x t . are
x + y x* + l
examples of rational functions.
* Zero exponents will be defined in Chapter 4. For now, one needs only note
that u = 1 for any non-zero number u.
62 The Function Concept Sec. 2-7
A number is an algebraic number if it is a root of a polynomial
equation of the form
a,QX n -{- aix n ~~* H~ 4~ a n ix -}- a n = 0,
in which the coefficients Oo, 0i, a n are integers, not all zero.
Analogous to the term algebraic number, we have the term
algebraic function. A function y = f(x) is called an algebraic func-
tion of x if y is a solution of an equation of the form
Po(x)y n + Pi(x)y*~ l + - - + P n ~i(x)y + P n (x) = 0,
where the coefficients PQ(X), PI(X), - - , P n (x) are polynomials in
x, and n is a positive integer.
Polynomials and rational functions are special types of algebraic
functions. The functions that we have considered so far were illus-
trations of algebraic functions. According to our definition,
y = \/# is an algebraic function of x because y 2 - x = 0. In this
case, ft = 2, P (a?)=l, Pi(x)=0, and P 2 (#) = x. Similarly,
y = y x "" i s an algebraic function of & because #7/ 2 - # 2 + 1 = 0.
Here n = 2, P (a;) = x, PI(X) = 0, and P 2 (z) = - a: 2 + 1.
An irrational function is an algebraic function which is not a
rational function.
A transcendental number is a number which is not algebraic,
and a transcendental function is a function which is not algebraic.
Functions like the trigonometric functions, which we shall take
up in Section 3-2, belong to the class of transcendental functions.
Later we shall consider other types of transcendental functions,
namely, the logarithmic and exponential functions such as log x
and
w Tfie Trigonometric Functions
3-1. THE POINT FUNCTION P(t)
The trigonometric functions that we are about to define are func-
tions in the sense previously described in Section 2-3 ; that is, they
are relations between two sets of numbers. The student who is
familiar with the trigonometric functions from his high-school work
is cautioned to note that we are not, for the present, discussing angles
in connection with these functions. We shall see that the concept
of a trigonometric function need not be associated with an angle;
in fact, many of the most important applications of mathematics
in modern science and engineering are concerned with trigonomet-
ric functions of pure numbers. Hence, we shall adopt the numer-
ical point of view, leaving the study in terms of angles as a
secondary consideration.
Consider a circle with a radius of one unit placed at the origin
of a rectangular-coordinate system. See Fig. 3-1. Let t be any real
ffi)
FIG. 3-1.
D
FIG. 3-2.
number. Starting at the point with coordinates (1,0), we lay off
on the unit circle an arc of length \t\. If t> 0, we measure the arc
in a counterclockwise direction. If t < 0, we measure in the clock-
wise direction. If t = 0, the arc consists only of the point (1,0). By
this procedure, there is associated with each real number t a defi-
nite end-point P(t) of the arc whose initial point is (1, 0). There-
63
64 The Trigonometric Functions Sec. 31
fore, corresponding to every real number t, we have a definite
ordered pair (x, y) of numbers which are the coordinates of the
endpoint of the arc.
Since P(t) lies on the unit circle, it is at one unit distance from
the origin. Hence, it follows from the distance formula that
(3-1) x 2 + y 2 = 1.
By means of this equation we can find the second coordinate of the
point P(t), except for sign, if one of the coordinates is known.
To determine the number pair (x, y) for the point P(t) corre-
sponding to a given value of t, we shall take note of the fact that the
circumference of the unit circle is 2?r = 6.2832 units (approxi-
mately) . For example, since arc ABC in Fig. 3-2 is one-half of the
complete circumference, it is TT units in length, and arc AB is equal
to 7T/2. It now becomes apparent that P(0) is the initial point
(1, 0) ; P(TT) is the point (-1, 0) ; P(?r/2) is the point (0, 1) ; and
P(37r/2) and P(-ir/2) both represent the point (0, -1).
3-2. DEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS
The correspondence between the set of real numbers t and the set
of ordered pairs (x, y) leads to definitions of the six common
trigonometric functions, namely, the sine, cosine, tangent, cotan-
gent, secant and cosecant.
General Relationships. We shall define the cosine of the real
number t to be x, and the sine of the real number t to be y. Thus,
we have
(3-2) x = cos t,
and
(3-3) y = sin t.
The domain of each of these functions (the sine function and the
cosine function) is the set of all real numbers t. Since, however,
every point P(t) lies on the unit circle, neither of its coordinates
(x, y) can exceed 1 in absolute value. Therefore,
(3-4) |cosJ|<;i and |sin|<;i.
In other words, the ranges of cos t and sin t are restricted by the
requirements 1 ^ cos t ^ 1 and 1 ^ sin t ^ 1, respectively, for all
values of t. It may be shown that the range of each of these func-
tions is the set of all real numbers u such that 1 = u ^ 1.
The other four functions may be defined in terms of the cosine
and sine, as follows :
(3-5) tan t = ^ (cos * * 0),
cos t
See. 3-2 The Trigonometric Functions 65
(3-6) cot t = 55i| (sin t * 0),
sin t
(3-7) sec t = -^- (cos < 5* 0),
cos t
(3-8) esc t = -r (sin * 5^ 0).
Since the cosine and sine are defined in terms of the coordinates
of the point P(t), it is also possible to express the other functions
in terms of these same coordinates. From the definitions of the
cosine and the sine given by (3-2) and (3-3) and from the defini-
tions of the other functions given by (3-5), (3-6), (3-7), and
(3-8), we have *
(3-9) ^ i =~ t =l' <"->
(3-10) <" = = ^ <""'
(3-11) sec*=-J = - G*?*0),
v ' cos i x - \ />
(3-12) esc t = -4- = - ' (y ^ 0).
sin t y
We note here that cos t, or x, appears in the denominators of both
tan t and sec t. Hence, tan t and sec t are not defined when t is a
number for which the ^-coordinate of P(t) equals zero. For
example, since the ^-coordinate of P(7r/2) or P(37r/2) is 0, it fol-
lows that tan 7r/2, sec 7r/2, tan 3?r/2, and sec 377/2 are not defined.
Similarly, it can be shown that cot 0, esc 0, cot TT, and esc TT are not
defined. We conclude, therefore, that the domain of each of the
functions tan t, cot t, sec t, and esc t is the set of all real numbers
for which the denominator is not zero.
It also follows from (3-7) and (3-8) that numerical values of
sec t or of esc t can never be less than 1. Hence, the ranges of
these two functions are restricted by the requirements
(3-13) |sec t\ 1 and |csc *| ^ 1.
From (3-5) and (3-6) we obtain an idea of the behavior of the
tangent and cotangent functions. It may be shown that the range
of each of these funptions is the set of all real numbers.
The Trigonometric Functions of ?r/6, Tr/4, and Tr/3. The computa-
tion of the numerical values of the trigonometric functions in gen-
eral is beyond the scope of this book. However, we shall use the
methods of plane geometry to find sin t, cos t, and tan t for ,= 7r/6,
66
The Trigonometric Functions
Sec. 3-2
FIG. 3-3.
FIG. 3-4.
FIG. 3-5.
77/4, and 77/8, in order to show that for certain values of t the
trigonometric functions can be found exactly without tables.
To compute the functions for the real value t 77/8, we con-
struct the unit circle of Fig. 3-3, Arc AB is given to be equal to
7T/3, which is one-sixth of the complete circumference. Triangle
OAB is inscribed in the circle, as shown, with side BO extended
through the origin to C. Our problem now is to find the values of
x = cos (7T/3) and y = sin (77/8) as coordinates of the point B.
Since OA = OB, triangle AOB is isosceles. Hence,
angle OAB = angle OBA.
We note also that arc CAB ~ 77 and
arc CA = arc CAB - arc AB = 77 - 77/8 = 277/3.
Therefore,
Also,
arc CA = 2 arc AB.
angle CO A = 2 angle AOB.
Furthermore, angle CO A is an exterior angle of triangle AOB.
Hence, it equals the sum of the two remote interior angles OAB
and OBA 9 or
angle COA = angle OAB + angle OBA.
Thus,
2 angle AOB = angle OAB + angle OBA,
and triangle AOB is equilateral.
If we draw BD perpendicular to OA, it will bisect OA. Then
x = 1/2 ; and from x 2 + y 2 = 1 it follows that y 2 ~ 3/4 and y = V5/2.
Hence, cos (77/8) =1/2, sin (77/8) = V5/2, and tan (77/8) = \/3 .
For i = 77/6, place the equilateral triangle AOB of Fig. 3-3 in the
unit circle as shown in Fig. 3-4, where E is the mid-point of arc
AB. Then arc EB = 77/6 and OE is the perpendicular bisector of
chord AB.
Sec. 3-2 The Trigonometric Functions 67
Since DB is one4ialf of chord^AB, y = 1/2. From x* + y 2 = 1 it
follows that x 2 = 3/4 and x = VS/2. Hence, cos (TT/G) = \/5/2 and
sin (7T/6) = 1/2. We then have tan (77/6) = 1/V3 = yff/3.
For = 7T/4, construct a unit circle as shown in Fig. 3-5 with
arc AB = Tr/4. Since arc AC = 7r/2, arc 5C = ir/2 - Tr/4 = ir/4.
Hence, arc AB = arc BC, and
angle A 05 = angle BOC.
Draw BZ) perpendicular to OA. Since the two parallels OC and
are cut by the transversal OB,
angle BOC = angle
Hence,
angle A0# = angle OBD,
and
Thus, y = x. Substituting this value of y in x 2 + y 2 = 1, we have
2x 2 = 1 and x 2 = 1/2. Therefore, x - cos (Tr/4) = \/2/2, and # =
sin (77/4) = V2/2. It follows-that tan (ir/4) = 1.
Other Special Values. In a similar fashion we can compute
rt_ C/jf
exactly the trigonometric functions of such values of t as > -5- >
7 Q 00
> and - -~ Functions of multiples of ?r/2 may also be computed
in this fashion, if one considers a straight line as a right triangle
in which one angle is and, hence, one side has zero length.
Example 3-1. Calculate the values of the six trigonometric functions of t = ?r/2.
Solution: As explained in Section 3-1, the coordinates of the point P(ir/2) are
(0, 1). Hence, by (3-2), (3-3), (3-9), (3-10), (3-11), and (3-12),
cos (7T/2) = 0, sec (?r/2) is undefined,
sin (?r/2) = 1, esc (7T/2) = 1,
tan (7T/2) is undefined, cot (w/2) = 0.
EXERCISE 3-1
1. Determine the coordinates of each of the following points:
a. P(2ir). b. P(-T). c. P(5x/2;.
d. P( - y) e. P(4ir). f. P(- 7ir).
2. In each of the following cases, carefully draw a unit circle and estimate the
coordinates of P(t).
a. P(l). b. P(2). c. P(3).
d. P(-2). e. P(4). f.-P(-8).
68
The Trigonometric Functions
Sec. 3-2
3. Evaluate each of the following:
d. cot-
g. csc(-).
K
b. cos -r-
o
. llTT
e. sin --
u / 7r \
h. sm (- ).
,
f. secy
4. In each of the following cases assume that f 2ir, and draw a figure showing
approximately the appropriate arc (or arcs).
a. sin t = 1/2. b. cot t = - 1. c. tan t = 1. d. esc t = - 1.
e. cos t = - 1/2 t sin t being positive. f. cot t = 1, sec i being negative.
5. Complete the following table, which shows the algebraic signs of the trig-
onometric functions in the four quadrants.
Quadrant in which P(t) lies
Function
I
II
Ill
IV
cos
+
"V
sin
+
+
-~
tan t
+
-
+
coU
sec
CSC
6. Use the equation x 2 + y 2 = 1 to find all values of t for which tan
where ^ J g 27r.
7. Use the equation x 2 + 1/ 2 = 1 in each of the following cases to find
trigonometric functions of t.
a. sin t = 1/2. b. cos = 3/5. c. sec J
d. esc t = - 3/2. e. sec t = - 2. f. tan t
g. cot t = 2. h. tan f = - 6/7. i. sin J
8. Prove that each of the following equations is correct.
a. sin ( = - sin t. b. cos ( i)
c. sec ( t) = sec . '^ d. tan ( t)
9. For each of the following cases, state the quadrant, or quadrants, in
given condition is satisfied.
a. The sine and cosine have the same signs.
b. The tangent and cosine have opposite signs.
t = cot t,
the other
= 13/12.
= 4/3.
= - 3/5.
cos f .
tan .
which the
3-3. IDENTITIES
As an immediate consequence of the definitions of the six trigo-
nometric functions, we can establish certain relationships among
Sec. 3-3 The Trigonometric Functions 69
them which hold for every value of t. Since P(t) lies on the unit
circle, (3-1) holds ; that is, x 2 + y 2 = 1. But, according to (3-2) and
(3-3) , x = cos t and y = sin t. Therefore, we have the equation
(3-14) cos 2 t + sin 2 t = 1.
This states that "the square of the cosine of t plus the square of
the sine of t equals unity." Since (3-14) holds for every value of t,
it is an identity. Note that we use the symbol cos 2 t instead of
(cos t) 2 . This simplified notation is used for all positive exponents,
but is never used in the case of a negative exponent. Thus, cos' 1 t
does not mean the same as (cos t)~ l , as we shall see in Chapter 8.
Similarly, we can prove that for each value of t for which the
functions are defined,
(3-15) 1 + tan 2 t = sec 2 ,
(3-16) 1 + cot 2 t = esc 2 t.
Proof of (3-15): By definition, tan t = ^ and sec t =
cos t cos t
However, these relationships have no meaning if cos t equals zero,
that is, if the ^-coordinate T)f P(t) equals zero. When cos 1 7*0,
we may divide both sides of the identity cos 2 1 + sin 2 1 = 1 by cos 2 *
and obtain
j sin 2 * 1
, c cos 2 t ~~~ cos 2 t
Iherefore,
1 + tan 2 t = sec 2 t.
nf\o f 1
Proof of (3-16): By definition, cot t = ~~ and esc t =
sin t sin t
In this case we assume that sin t ^ 0. When we divide both sides
of sin 2 1 + cos 2 1 1 by sin 2 1, we obtain
cos 2 t _ 1
sin 2 t ~~ sin 2 t
Therefore,
1 + cot 2 t = esc 2 *.
We thus have established the three identities which we restate
here for easier reference :
(3-14) cos 2 t + sin 2 1 = 1,
(3-15) 1 + tan 2 t = sec 2 t,
(3-16) 1 + cot 2 t = esc 2 t.
These fundamental! identities are very important in working with
trigonometric identities and should be remembered.
Our present work with identities will consist of reducing given
trigonometric expressions to other forms. Unfortunately, no specific
rule of procedure can be given for making these reductions. Profit-
70 The Trigonometric Functions Sec. 3-3
ciency in making such reductions is a matter of both practice and
experience. Generally speaking, when we want to reduce a given
expression to some other form, it is helpful first to perform any
indicated algebraic operations and then to use some form of one of
the fundamental relationships to simplify the expression.
To prove an identity, we may proceed in any one of the follow-
ing ways :
1) We may work on the more complicated member of the identity
and attempt to reduce it to the simpler member.
2) We may work with both sides of the identity and show that they
induce to the same expression.
3) We may form the difference of the two sides of the identity and
prove that difference to be equal to zero.
It is frequently desirable to express both sides of the given identity
in terms of sines and cosines alone, and then use (3-14) if needed.
We shall consider a few examples to illustrate the procedure in
reducing expressions.
Example 3-2. Show that cos t -f sin t tan t = sec t.
Solution: From (3-5), or the definition of the tangent, we have
cos t -f sin t tan t = cos t -f sin t
.,,. , cos t
Adding, we have t cog2 , ^ 2J
cos t -f sin t =
cos t cos t
Since cos 2 1 + sin 2 1 = 1 and = sec t,
cos t
cos t + sin t tan t = : = sec L
cos t
Example 3-3. Reduce - to sin t cos t.
* tan t + cot t
Solution: By definition, tan t = and cot t = - - Therefore,
J ' cost sin t
111 cos t sin t
tan t + cot t sin t cost sin 2 1 + cos 2 1 sin 2 t + cos 2 1
cos t sin t cos t sin t
Since sin 2 1 + cos 2 1 = 1, we obtain
1 cos t sin t
tan t + cot t
= sin t cos t.
Example 3-4. Establish the identity (sec t - cos O 2 = tan 2 t(l - cos 2 t) by
reducing both sides to the same expression.
Solution: By definition, sec t = Hence, the left side becomes
cos t
f I .\* /I - cos 2 1\*
[ T - COS = I 1 )
Vcos t / \ cos t )
Sec. 3-4 The Trigonometric Functions 71
By (3-14) we have
/I cos 2 *\ 2 _ /sin 2 *\ 2 _ sin 4
\ cos t ) ~~ Vcos t) "" cos 2
cos 2 1
The right side of the given identity may be reduced as follows. By definition,
sin t
tan t =
Hence, we have . _ . . , .
x o 1/1 o i\ sin 2 1 . . . sin 4 t
tan 2 J(l - cos 2 1) = sin 2 1 = ~-
<?os 2 cos 2 1
Since both sides reduce to the same expression, > the identity is established.
cos t
EXERCISE 3-2
Prove each of the following identities:
- . . ~ rt sin , cos t -
1. sin - cos t tan = 0. 2. H = 1.
esc t sec t
sin t sec ~ A , . . .
3. : ; = 0. 4. tan J esc t = sec f.
cos esc
5. sin (cot t + esc = 1 H- cos t. 6. (sin cos t) 2 = 1 2 sin cos J.
7. sec 2 + 2 tan t = (1 + tan t) 2 . 8. sin (csc J - sin t) = cos 2 *.
9. tan 2 J(cot 2 - cos 2 = cos 2 L- 10. cos 2 J -I- cos 2 1 tan 2 J = 1.
11. (1 - sin (1 + sin ) = cos 2 1. 12. (sec J - 1) (sec t + 1) = tan 2 J.
13. -~ = esc ^. 14. sin t tan -f cos 2 J sec i = sec t.
I - cos 2 1
15. sin 2 esc 2 1 = 2 - cos 2 sec 2 . 16. sin 4 1 sec 2 J esc 2 1 = tan 2 J.
17. tan t + cot = sec J esc J. 18. esc 2 1 + sec 2 1 = sec 2 esc 2 J.
19. sec 4 - sec 2 1 = tan 4 J + tan 2 J. 20. sin 4 1 - cos 4 J = sin 2 1 - cos 2 .
21. sm * tan * = . S ec t. 22. tan 2 < + cot 2 1 = sec 2 csc ? ^ - 2.
cos 2 1 1
2^ 1 ^ sm ^ , 1 - cos t __ sin t + cos t 1 ^
cos t sin ~~ sin t cos ^
Ojl sec 2 t - tan 2 + tan t ... .
24. ; = sin t 4- cos t.
sec t
25. sec 2 1 = esc 2 <(sec 8 - 1). 26. (tan * + cot O 2 = sec 2 1 esc 2 .
27. 1 -f tan 2 < = (sec 2 1 - l)csc 2 i. 28. sin t(l + tan 2 - sin < = tan 3 1 cos J.
29. rinf-coBJ tan|-l ^ sin3 1+ cosM = x ^ ^ f ^ ^
sin ^ + cos t tan + 1 sin t -f cos i
1 " f*os ^
31. cot 2 1 - cos 2 J = cos 2 1 cot 2 . 32. : . . = (esc t - cot O 2
1 + cos i
3-4. TABLES OF TRIGONOMETRIC FUNCTIONS
Exact numerical values of trigonometric functions in general
cannot be found. However, by use of methods beyond the scope of
this book, the values can be computed to as many decimal places as
desired. The results of such computations have been tabulated an<l
are included in this text in the form of tables of trigonometric
functions. Table I at the end of the text contains the values of the
The Tr/gonomefr/c Funcfions
Sec. 3-4
P(t)
FIG. 3-6.
six functions corresponding to numbers t such that ^ t ^ ir/2.
Actually, since ir/2 = 1.5708 approximately, the table contains
values of between and 1.60.
Let P(t) (x, y) be the point corresponding to a given value of
t, Fig. 3-6, and let ti denote the length of the shorter arc which
joins P(t) to the o>axis. In each case in Fig. 3-6, the point P(ti)
is located by measuring the arc t\ counterclockwise from the posi-
tive half of the #-axis. We shall call ti the reference number, or
related number, for t. Note that ^ t\ ^ rr/2. Since t v is a real
number, there is associated with it a point P(t\) = (#1, 2/1). Also,
since ti lies between and 77/2, P(ti) must lie in the first quadrant.
In each case in Fig. 3-6 the coordinates of P(t^) must be numer-
ically equal to those of P(t) ; that is, \x\ = Xi and \y\ = y\. Since
all the trigonometric functions are defined in terms of x and y, we
can say that
(3-17) | any function of t j = same function of t\.
These functional values may not have the same algebraic sign,
since P(ti) lies in the first quadrant and all functions of t\ have
positive values, whereas P(t) may lie in any quadrant and the
functions of t do not necessarily have positive values. It is impor-
tant to see that the proper sign is chosen to make the equation a
true one. The algebraic sign in each case depends on the quadrant
in which P(t) lies.
The following examples and Fig. 3-7 will illustrate the method
of reducing a function of any positive or negative t to the same
f unctipn of the reference number ti.
We shall limit our discussion for the present to direct use of
Table I and consider ipnly values of ti which are shown there. The
process of using the table for values of ti which are not shown will
be treated in Section 3-10 when we discuss interpolation. For
simplicity at this time, we shall use the approximate value TT = 3.14.
Sec. 3-4
The Trigonomefric Functions
73
Example 3-5. Reduce the functions of t = 2 to functions of its reference number.
Solution: As shown in Fig. 3-7(a), P(2) is in the second quadrant, and the
reference number t\ is TT 2, or 1.14. The numerical values of the functions
of 1.14 may be found from Table I. The signs of the functions of 2 are determined
by noting that only the sine and cosecant are positive in the second quadrant.
Thus, we have :
cos 2 = - cos (T - 2) = - cos 1.14 = - 0.4176,
sin 2 = sin (w - 2) = sin 1.14 = 0.9086,
tan 2 = - tan (w - 2) = - tan 1.14 = - 2.176,
cot 2 = - cot (T - 2) = - cot 1.14 = - 0.4596,
sec 2 = - sec (T - 2) = - sec 1.14 = - 2.395,
esc 2 = esc (TT - 2) = esc 1.14 = 1.101.
Example 3-6. Find tan 4.
Solution: As shown in Fig. 3-7(6), the reference number t\ is 4 ir = .86.
Since the tangent is positive in the third quadrant, tan 4 = tan .86 = 1.162.
Example 3-7. Find cos 5.
Solution: Here, as shown in Fig. 3-7(c), t\ = 2?r 5 = 1.28; and cos 5 =
cos 1.28 = 0.2867.
Example 3-8. Find cot 20.
Solution: To locate the point P(20), we must proceed 20 units around the
circle in a positive direction from (1, 0). The number of units in one complete
revolution is 2ir = 6.28, find we find that
20 = 3(6.28) + 1.16.
Therefore, to locate P(20), we must proceed three times around the unit circle and
then continue for 1.16 additional units in a counterclockwise direction. This means
that t may be taken as 1.16. Since P(20) or P(1.16) lies in the first quadrant,
*! = t = 1.16. From the table, we have cot 20 = cot 1.16 = 0.4356.
74
The Trigonometric Funcf/ons
Sec. 3-4
Example 3-9. Find sin (-2).
Solution: For t = - 2, it is shown in Fig. 3-8 that i = |- * - (- 2)|
= | - IT -f 2| = 1.14. Hence, t\ is the same as t\ in Example 3-5. Since sin t is
negative in the third quadrant, we have
sin (- 2) = - sin 2 = - sin 1.14 = - 0.9086.
P(2)
It should be noted that in Fig. 3-8 the
points P(-t) andP(t) are located on oppo-
site sides of the #-axis, and are joined by a
line segment which is bisected perpendicu-
larly by the axis. Hence the coordinates of
the two points are numerically equal, but
the ^-coordinates have opposite signs.
Therefore, by the definitions of the func-
tions given in Section 3-2, it follows that
sin ( ) = sin t,
cos ( t) = cos t,
tan ( = -tan t,
esc (t) = esc t,
sec (t) = sec t,
cot ( t) = cot t.
Hence, if t > 0,
(3-18) (any function of (-t) | = |same function of t\.
However, the algebraic sign of the function is changed for all
functions except the cosine and the secant.
Because cos t and sec t remain unchanged when t is replaced by
its negative, these functions are called even functions. The remain-
ing functions are called odd functions, since their values change
sign when t is replaced by its negative.
EXERCISE 3-3
1. Construct a figure, locating each of the following points. Show the point P(t)
and its related number ti. (Use TT = 3.14).
a. P(l). b. P(3). c. P(10). d. P(- 5). e. P(- 4). f. P(3/2).
2. With the aid of Table I find each of the following values, using TT = 3.14.
a. sin 1.45. b. cos 3.5. c. sec 4.75.
d. tan 5. e. esc (- 2.41). f. cot (- 4.50).
g. sin 28. h. cos 60. i. tan ( - 30).
j. sec
Sir
k. cot
4
1. sin
227T
3-5. POSITIVE AND NEGATIVE ANGLES AND STANDARD POSITION
We have defined each of the six trigonometric functions as a
relation between two sets of numbers, employing as the independent
Sec. 3-5
The Trigonometric Functions
FIG. 3-9.
variable a real number
t whose absolute value
represents the length
of an arc of a unit
circle. Now we shall
return to the tradi-
tional viewpoint and
consider trigonometric
functions of angles.
Although the stu-
dent is probably famil-
iar with the idea of
angle from the study of geometry, we shall try to make the definition
more precise. Let us select a point in a plane and draw the half-
line or ray a emanating from O, as shown in Fig. 3-9. We shall
call the vertex of the ray. Finally, we let A be a point on the ray
in its initial position.
Now rotate the ray a about^O to some terminal position 6, so that
the point A moves along the arc indicated by the curved arrow AB.
The ray may be rotated in the counterclockwise sense, as in Fig.
3-9 (a) , or in the clockwise sense, as in view (b). Moreover, it may
be turned through one or more complete revolutions, as in view (c) .
We shall speak of the position b as the terminal ray b. We have
then an ordered pair of half -lines consisting of the initial ray a and
the terminal ray b. We can now define an angle as follows :
An angle is a geometric figure consisting of two ordered rays
emanating from a common vertex.
With each angle is associated a number, called the measure of the
angle, which indicates the sense and amount of rotation required
to turn from the initial ray of the angle to the terminal ray. This
rotation is usually represented graphically by a curved arrow. Its
evaluation will be considered in Section 3-6.
We may designate the angle in Fig. 3-9 as angle AOB; or we may
use a Greek letter, such as 0, $, a, /3, or y, as the designation. The
line OA is called the initial side of angle AOB f and OB is the
terminal side. Counterclockwise rotation, as in Fig. 3-9 (a) 01
3-9 (c), gives rise tb a positive angle, while clockwise rotation,
such as the one in Fig. 3-9(6), gives rise to a negative angle.
Finally, we shall say that an angle is in standard position with
respect to a rectangular coordinate system when its vertex is at the
origin and its initial side coincides with the positive #-axis. See
76
The Trigonometric Functions
Sec. 3-5
III
FIG. 3-10.
FIG. 8-11.
Fig. 3-10. When an angle is placed in standard position, the
terminal side determines the quadrant to which an angle is said
to belong. Thus, angle XOP in Fig. 3-10 is positive because it is
generated in a counterclockwise direction, and is a second-quadrant
angle because the terminal side OP lies in the second quadrant.
We note that the definition of angle does not specify that the
rotation should stop at the first arrival at the terminal side OP,
Fig. 3-10. In fact, angles of any size may be generated, since any
number of angles which end at the terminal side OP of a given
angle may be obtained simply by adding a number of complete
rotations, positive or negative, to the given angle. For example, the
same terminal side may also be reached by rotation in the opposite
direction. All angles which are in standard position and have the
same terminal sides are called coterminal angles.
In Fig. 3-11 the angle a is generated by rotation of OX counter-
clockwise to the position OP. The angle /3, which is coterminal with
a, is generated by adding to a one complete rotation of OX. The
angle y is a negative angle, which is coterminal with a and is
generated by rotating OX in the clockwise direction to the
position OP.
3-6. MEASUREMENT OF ANGLES
The problem of measuring an angle is
equivalent to that of finding the measure
of the associated arc. One should, there-
fore, apply the discussion of Section 3-1
and construct a unit circle as shown in
Fig. 3-12.
Let be an angle in standard position.
Since the initial and terminal sides of the
angle intersect the circle in the points P (0)
FIG. 3-12. an( j P(t), respectively, the problem of
measuring the angle reduces to that of measuring the appropriate
$ec. 37 The Trigonometric Functions 77
arc length t. Thus, the measure of the angle can be found in terms
of a real number in any one of several ways, depending on the unit
of measure chosen.
We shall consider first the circular system, or natural system, of
measuring angles, which is used almost exclusively in the calculus
and its applications. Its fundamental unit is the radian. This unit
may be defined as follows :
A radian is the measure of an angle which, if placed at the center
of a circle, intercepts an arc on the circumference equal in length
to the radius of the circle.
In Fig. 3-13 the angle AOB is
1 radian, and the length of the
subtended arc AB is equal to the
radius r. / \ arc =: radius
If the circle selected for meas-
uring a radian is a unit circle, we
have an alternate definition of a m
radian. That is, a radian is an o radius =r A
angle which intercepts a unit arc F IG . 3.43.
on a unit circle.
Another system of measuring angles is the sexagesimal system,
or degree system, which is commonly used in ordinary calculations
involving angles. The fundamental unit of this system is the degree.
In Section 3-7 we shall study various relations between radians
and degrees, and shall develop rules which allow us to convert
from one system to the other.
In discussing angles, we frequently use the term angle, in place
of measure of an angle, and we rely on the context to make the
meaning clear. Thus, when we say "0 = 2," we mean, "0 is an angle
whose measure is 2 radians." The word radian is usually omitted
when an angle is expressed in terms of radians.
3-7. THE RELATION BETWEEN RADIANS AND DEGREES
Since an arc that is equal in length to the radius of a circle sub-
tends an angle of one radian at the center, it follows that the whold
circumference, which is 2rr times the radius, subtends an angle of
2?r radians. Furthermore, the whole circumference subtends a
central angle of 360. Therefore,
2?r radians = 360,
and
TT radians = 180.
78 The Trigonometric Functions Sec. 3-7
If the approximate value 3.1416 is used for TT,
180 180
1 radian = - = Q 1 . 1A (approximately),
7T U.141D
or
1 radian = 57.29578 (approximately),
or
1 radian = 5717'45" (approximately).
Also,
1 = T^T radians = 0.01745329 radians (approximately).
In order to make the conversion to radians easier when the angle
is expressed in degrees, minutes, and seconds, we give the following
values :
1' = 0.00029089 radians,
and
1" = 0.00000485 radians.
Therefore, one of the following rules can be used to convert from
degrees to radians or from radians to degrees :
To convert from degrees to radians, multiply the number of
degrees by ~ , or 0.0174533.
ioU
To convert from radians to degrees, multiply the number of
radians by , or 57.29578.
7T
Note, However, that certain angles are commonly expressed in
terms of TT radians, in order to avoid approximate values. For
example,
180 = TT radians, 45 = 7T/4 radians,
90 = 7T/2 radians, 30 = 7T/6 radians.
3-8. ARC LENGTH AND AREA OF A SECTOR
In Fig. 3-14 is shown a circle of radius r. In such a circle an
angle at the center equal to one radian subtends an arc on the
circumference equal to r. Similarly, by the definition of a radian,
the number of units in the arc s intercepted by a central angle equal
to radians is given by the relationship
Thus,
(3-19) $ = re,
or
FIG. 3-14.
arc = (radius) (central angle expressed in radians) .
Sec. 3-8 The Trigonometric Functions 79
Now let A denote the area of the sector bounded by two radii
and an arc of length s. If is the number of radians in the central
angle of the sector, then the ratio of the area A of the sector to the
area of the whole circle, or Trr 2 , equals the ratio of the angle to
the angle in the whole circle, or 27r. That is,
A = A
Trr 2 2?r '
or
(3-20) A =$'* 9 '
If the central angle of an arc or a sector is expressed in degrees,
it must be re-expressed in radians before (3-19) or (3-20) can
be applied.
Example 3-10. Express 210 in terms of TT radians.
Solution: Since 1 =
lo
Thus, 210 = - radians.
Solution: Since 1 = ^ radians, 210 = 210 ~ = ^ radians.
loU loU O
Example 3-11. Express 1215'20" in radians.
Solution: Multiply the decimal parts of a radian given in Section 3-7 for 1, 1',
and I" by 12, 15, and 20, respectively. The results are as follows:
12 0' 0" = .20943948 radians
15' 0" = .00436335 radians
20" = .00009700 radians
12 15' 20" = .21389983 radians.
Example 3-12. Express -^- radians in degrees.
Solution: Since TT radians = 180, ^ radians = | (180) = 150.
6 o
Example 3-13. Express 3.5 radians in degrees, minutes, and seconds.
Solution: First, convert the radians to degrees, as follows:
3.5 radians = (3.5) (57.29578) = 200.5352.
To find the number of minutes, multiply the decimal part of a degree, or 0.5352,
by 60. Thus, 0.5352 = (60) (0.5352) minutes = 32.112'.
To find thfe number of seconds, multiply the decimal part of a minute, or 0.112,
by 60. The result is 0.112' = (60) (0.112) seconds = 6.72*.
Hence, 3.5 radians = 20032'6.72".
80 The Trigonomefric Functions Sec. 3-8.
Example 3-14. The radius of a circle is 5 inches. Find the length of the arc of
the circle subtended by a central angle of 30.
Solution: Since 30 = -^ > the central angle is ~- Also, r = 5. Therefore,
by (3-19),
=r-0=5--=j| (3.1416) = 2.618 inches,
o o
Example 3-15. In a circle of radius 6 inches, what is the area of a sector whose
central angle is 60?
Solution: By (3-20), the area of the sector is ^r 2 0. Since 6 = 60 = ~ ,
& 3
A = 0(36)7,- = 6?r square inches.
t O
EXERCISE 3-4
In each problem from 1 to 25, express the given angle in radians.
1. 60. 2. 45. 3. 30. 4. 10. 5. 120.
6. 150. 7. 12. 8. 90. 9. 240. 10. 330.
11. 72. 12. 20. 13. 215. 14. 196. 15. 321.
16. 283. 17. 63. 18. 3010'. 19. 4621 / . 20. 23637'.
21. 8216'. 22. 6321'17". 23. 18357'43". 24. 39244'27". 25. 9331'38*.
In each problem from 26 to 40, express the given angle in degrees.
26. 7T/6. 27. 7T/4. 28. Tr/8. 29. 37T/2. 30. 47T/5.
31. T/12. 32. Sir/18. 33. 7<jr/2. 34. 5ir/3. 35. 3?r/20.
36. 3.7 rad. 37. 8.21 rad. 38. 0.34 rad. 39. 0.763 rad. 40. 0.8136 rad.
In each problem from 41 to 56, draw the given angle in standard position and
indicate its terminal side.
41. 30. o 42. 7T/4. 43. ir/3. 44. 90.
45. 22^- 46. 27T/3. 47. 170. 48. 17T/18.
49. - T/2. 50. 630. 51. 360. 52. - 47T.
53. 77T/3. 54. 1000. 55. 97r/4. 56. - llTT/6.
57. In a circle of radius 4 feet, find the length of the arc intercepted by an angle of
7ir/6 radians. Find the angle in radians that intercepts a 5-foot arc.
58. A central angle in a circle of radius 15 inches intercepts an arc of 5 inches.
Find the number of radians in the central angle. Express this angle in degrees
and minutes, rounding off the result to the nearest minute.
59. A central angle of 6214' intercepts an arc of 16 inches on the circumference of
a circle. Find the radius of the circle.
60. Find the area of a circular sector whose radius is 7 inches and whose central
angle is o) 4 radians; 6) 75; c) 3 radians.
61. The area of a circular sector is 72 square inches. Find the angle if the radius
is a) 6 inches; 6) 9 inches; c) 5 feet.
62. The area of a circular sector is 126 square inches. Find the radius if the anglfe
is a) 128; 6) 1.6 radians; c) 30.
Sec. 3-9 The Trigonometric Functions 81
3-9. TRIGONOMETRIC FUNCTIONS OF ANGLES
Let be an angle in standard position, as shown in Fig. 3-15.
With 6 we can associate a real number t, which is the measure of
the angle in radians. This concept is equivalent to our previous
concept of t, when t was interpreted as the length of an arc laid off
on the unit circle by starting at the point (1, 0) and terminating at
P(*,y)
FIG. 3-15.
FIG. 3-16.
the point P(t). Such an association of the angle 6 with the
directed length t of an arc of a unit circle allows us to define the
cosine and sine of as cos = cos t and sin = sin t. A similar
procedure may be followed for the other functions of 0.
Now consider Fig. 3-16, where we show an angle in standard
position and a unit circle. By the definition of 0, the terminal side
of intersects the unit circle at the point (cos 0, sin 0). This is, of
course, the point designated previously as P(t). We now extend
the terminal side of to an arbitrary point P with coordinates
(x, y). The length of the radius vector OP is r = V# 2 + 2/ 2 *
If we drop perpendiculars from the points (cos 0, sin 0) and
(x,y) to the re-axis, the right triangles thus constructed are similar.
Therefore,
x cos , y sin
- = : and - = T
r 1 r 1
Hence, the coordinates of the point P(x,y) on the terminal side are
x = r cos and y = r sin 0.
Using these results with the definitions of the functions from
Section 3-2, we caii express the values of the six functions in
terms of x, y, and r. Thus,
sin = 2//r, esc = r/y,
(3-21) cos = x/r, sec = r/x,
tan = y/Xj cot = x/y.
82 The Trigonometric Functions Sec. 3-10
3-10. TABLES OF NATURAL TRIGONOMETRIC FUNCTIONS OF ANGLES
Tables of natural trigonometric functions are so labeled to dis-
tinguish them from tables of the logarithms of these functions.
Angles in Radians* In Section 3-4 Table I was used to find values
of trigonometric functions of the type cos 2 or sin 27T/3. On the
basis of the definitions of the functions of an angle given in Section
3-9, Table I may also be used to find the functions of angles meas-
ured in radians.
Example 3-16. Find the cosine of an angle of 1.43 radians.
Solution: From Table I, cos 1.43 = 0.1403.
Angles in Degrees. Table II at the end of this text contains the
approximate values of the six functions of acute angles expressed
in degrees and minutes. It is a four-place table of the functions of
angles at intervals of 10 minutes.
To find the value of a function of an angle between and 45,
first locate the angle in one of the columns at the left, and then
look for the value on the same line in the column headed by the
name of the desired function. For an angle between 45 p and 90,
locate the angle in a column at the right, and then look for the
value on the same line in the column with the name of the desired
function at its foot.
Table II should be referred to in working through these illustra-
tive examples.
Example 3-17. Find sin 3240'.
Solution: This angle is between and 45. Look in the left-hand column to find
3240', and then go to the right to the column headed sin. There find 0.5398. Hence,
sin 3240' = 0.5398.
Example 3-18. Find cos 5620'.
Solution: This angle is between 45 and 90. So look in the right-hand column
to find 5620', noting that 5620' is above 5600', and then go to the left to the
column with cos at its foot. Thus, cos 5620' = 0.5544.
The following examples illustrate the procedure for finding an
angle corresponding to a given value of a function.
Sec. 3f-10 The Trigonometric Functions 83
Example 3-19. Given tan 6 = 4.511, find 6, assuming that ^ 90.
Solution: Since tan is greater than 1, is greater than 45. Therefore, search
through the columns marked tan at the foot for the given number 4.511. The
corresponding angle in the right-hand column is 7730'. So 4.511 = tan 7730',
or = 7730'.
Example 3-20. Given cos 6 = 0.8660, find 6, assuming that ^ ^ 90.
Solution: By looking through the columns with cos at either the head or the foot,
find 0,8660. Since this value is in a column headed cos, use the left-hand column for
the corresponding angle, which is 30. Hence, 6 = 30.
Interpolation. When either the given angle or the given value of
a function is not printed in the table, we can find the desired value
or angle by using a method of approximation known as interpola-
tion. We assume that the change in the value of the function is
directly proportional to the change in the angle. Although this
assumption is not strictly valid, it gives values that are accurate
enough for many practical purposes if we limit its use to small
changes in the angle.
The process of direct interpolation is used if the angle is given
and we need to find the value of either an increasing function of
the angle, such as the sine, or a decreasing function, such as the
cosine. Inverse interpolation is used when the value of a trigo-
nometric function is known and the angle is to be found.
Example 3-21. Find sin 1812'.
Solution: This angle is not listed in the table, but it lies between 18 9 10' and
1820'. From the table we find that
sin 1810' = 0.3118,
and
sin 1820' = 0.3145.
The desired value of sin 1812' will then lie between 0.3118 and 0.3145.
The tabular difference, that is, the difference between the two values listed in
the table, is 0.0027. Also, the difference between the angles 1810' and 1820' is
10', while the angle 1812' differs from 1810' by 2'. Since the change in the angle
from 1810' to 1812' is 2/10 of the change from 1810' to 1820', we assume that
the corresponding change in the value of the sine will be (0,2) (0.0027) = 0.0005,
and the amount to be added to 0.3118 is 0.0005. Hence, sin 1812' =0.3123.
The accompanying diagrammatic arrangement presents this same operation in
tabular form: f sin 18 o 10 ' = 0.3118 \
2 1*
10 [ sin 1812' = 0.3118 + x J 0.0027
sin 1820' = 0.3145
84 The Trigonometric Functions Sec. 3-10
Since the angle 1812' is 2/10 of the way from 1810' to 1820 7 , the corresponding
functional value will be 2/10 of the way from 0.3118 to 0.3145. Therefore,
0.0027 10
r x = (0.2) (0.0027) = 0.0005.
This amount is to be added to 0.3118. Hence, the value of the function is 0.3123,
and sin 1812' = 0.3123.
Example 3-22. Find cos 7348'.
Solution: The process is similar to that in Example 3-21. However, since the
cosine decreases as the angle increases, we subtract 8/10 of the tabular difference
from cos 7340'. We find the values of cos 7340' and cos 7350' in a column of the
table labeled cos at the bottom. The work may be indicated as follows:
10
cos 7340' = 0.2812
x
cos 7348' = 0.2812 - x
cos 7350' = 0.2784
x
0.0028
= , or x = (0.8) (0.0028) = 0.0022.
0.0028 10
Hence, the amount to be subtracted from 0.2812 is 0.0022, and cos 7348' = 0.2790,
The inverse process of finding the angle when the given value of
a function is not printed in the table is performed in a similar
fashion. Here, since we know the value of the function, we find
the two values in the table nearest the given value, one less than
it and one greater. Again making the assumption that small
changes in the value of the function are proportional to small
changes in the angle, we proceed as indicated in the following
example.
Example 3-23. Find if cot 6 = 0.8780.
Solution: This value of the cotangent is not in the table but lies between the
entries 0.8796 and 0.8744. To these correspond, respectively, the angles 4840'
and 4850 / . We have, therefore, the following tabulation:
10
cot 4840' = 0.8796
x { \ 0.0016
cot 48 (40 + xY = 0.8780
cot 4850' = 0.8744
0.0052
* 16 ,
To = W ' and x =:
Hence, 9 = 48*43'. 1U 5J
Sec. 3-10 The Trigonometric Functions 85
EXERCISE 3-5
In each of the problems from 1 to 30, use Table II to find the value of the given
function. Interpolate whenever necessary.
1. sin 3620'. 2. cot 12840'. 3. sec 2340'.
4. cos 9650'. 5. sin 13210'. 6. tan (- 2S10').
7. esc 22330'. 8. sec 3930 / . 9. cot 28350'.
10. sin 9840'. 11. cos 75CO'. 12. cot (- 13330').
13. sec (- 39210'). 14. esc (- 41620 / ). 15. tan 62340'.
16. tan 29852'. 17. cot 5543'. 18. esc (- 4451').
19. sin 5732'. 20. cot 31G'. 21. sin (- 28033').
22. cot 2801'. 23. tan 271G / . 24. esc (- 24529').
25. cos (- 7258'). 26. sin 31237 / . 27. tan 63602'.
28. sin (- 1647'). 29. esc 289OG'. * 30. cos 12619'.
In each of the problems from 31 to 60, use Table II to f.nd the values of between
and 360 which satisfy the given equation. Express the results to the nearest
minute, interpolating whenever necessary.
31. tan = - 0.11C8. 32. sin = 0.3062. 33. cot = 1.091.
34. cos = 0.7951. 35. tan = 0.0553. 36. sin = 0.2419.
37. sin = 0.5783. 38. cot = - O.G494. 39. tan = 1.511.
40. cos = - 0.4147. 41. shT0 = - 0.9959. 42. cot = 0.0437.
43. tan = 8.345. 44. cot = - 0.3121. 45. tan = 1.446.
46. sin = O.G702 47. tan = 0.9043. 48. cot = 2.398.
49. ccs = 0.9503. 50. cos = - 0.5090. 51. cos = 0.8519.
52. cot = - 1.381. 53. cot = 0.4230. 54. sin = 0.2491.
55. cot = 7.COO. 56. tan = - 0.1191. 57. cos = - 0.1323.
58. cot = 0.1340. 59. tan = - LS.CO. 60. tan = 3.235.
In each of the problems irom Gl to 72, find the value of the given function.
Interpolate whenever necessary. Take TT as 3.14.
61. sin 0.93. C?. cot 2.46. 63. sec (- 1.24).
64. tan 8.71. C5. esc 9.43. 66. cot 0.678.
67. tan 0.333. 68. cot ( - 1) 69. cos ^
70. sin ( - -?} 71. esc 0.968. 72. cot (- 0.643).
\ 6 /
In each of the problems from 73 to 84, find the values of 0, in radians, between
and 2ir \\hich satisfy the given equation. Use Table I and express the results to
three decimal places, interpolating whenever necessary.
73. cos = 0.9759. 74. sin = 0.9967. * 75. tan = 2.066.
76. sec = 2.563. 77. tan = 0.9413. 78. sin = - 0.736$
79. cos = 0.4010. 80. tan = 1.6. 81. sin = 0.91.
82. cot = 0.39. 83. cos = 0.84. 84. cot = 1.031.
4
The Laws of Exponents
4-1. POSITIVE INTEGRAL EXPONENTS
When studying the progress of algebra up to the sixteenth
century, one cannot help but be perplexed by either the total
absence of symbolism or, when present, the lack of uniformity in
its use. At first, unknown quantities were often represented by
words. Later, symbols made from abbreviations and initial letters
of these words were used to indicate mathematical concepts, such
as number, power, and square.
Descartes (1637) is generally credited with our present system
of exponents. He introduced the Hindu-Arabic numerals as expo-
nents, using the notations a, aa [sic] , a 3 , a 4 , etc. The writing of a
repeated letter for the second power of the unknown continued for
many years.
Laws for positive integral exponents were introduced in Section
1-11, without proofs. We shall now establish these laws and extend
them to apply also to zero, negative, and fractional exponents.
We recall that if n is any positive integer, a n means the product
of n factors each equal to a. In this notation, a is the base and n is
the exponent or power. We shall proceed to establish the following
laws for positive integral exponents.
Law of Multiplication. If a is a real number, and if ra and n are
positive integers,
(4-1) a m a n = a m + n .
Proof. Proof of this relationship follows from the definition of a n
and the associative law for multiplication. Thus,
a m = a a a (to m factors),
and
a n = a a a (to n factors).
Hence,
a m a n = [a a a(to m factors)] [a a a(to n factors)]
= a a cr(to m + n factors)
= a m+n .
For example, x 3 x 5 = a; 8 , and y k y k+3 = y 2k + 3 .
86
Sec. 4*1 The Laws of Exponents 87
Law of Division. If a is a non-zero real number, and if m and n
are positive integers such that m > n, then
(4-2) = a m ~ n .
\ f a n
If a T^ 0, and if n > m, then
n m 1
(4-3)
a n a n ~~ m
Proof. Proofs of these relationships follow :
If m > n, then m n is positive. By (4-1) ,
a m ~ n a n = 0(w n)+n = o, m .
Hence, dividing both sides by a n , we have
a m
a mn = - .
a n
For example,
3 7
~5 > 35 -
If m < n, then n m is positive. By (4-1),
a a n ~ = a w+(n ~ m) = a n .
Divide both sides by a n ~ m to obtain
a n
a m
a n-m
Now, dividing both sides by a n , we have
a m ( a n \ / a n 1 I
= ( - ) / a n = - =
a n \a n - m / / a n a n ~ m a n ~
For example,
_
37 - 32 '
Law for a Power of a Power. If a is a real number, and if m and
n are positive integers, then
(4-4) (a m ) n = a mn .
Proof. This relationship can be easily proved as follows :
By the associative laws for multiplication and addition, the law
of multiplication expressed by (4-1) can be extended to three or
more factors. Thus,
0/n . gn . QP rr (a m a n ) Q?
=. a m+n a p
A similar relationship can be written for any number of factors.
That is,
(a m ) (of) - (a r ) =
88 The Laws of Exponents Sec.
We may now take m = p = = r to get
(a m ) (a m ) (a m ) (to n factors) = a m+m + + = a 1 "*.
Hence,
(a w ) w = a mn .
For example,
(z 2 ) 3 = x 6 , and (2 2 ** 1 ) 5 = 2 10 *+ 5 .
Law for a Power of a Product. If a and 6 are real numbers,
and if m and n are positive integers, then
(4-5) (ab) n = a n b n .
Proof. In proving this relationship, we make use of the associa-
tive and commutative laws of multiplication. Thus,
(ab) n = (ab) (ab) ..... (ab) (to n factors)
= [a a ..... a(to n factors)] [6 b ..... b(to n factors)]
= a n b n .
For example,
Law for a Power of a Quotient. If a and 6 are real numbers, if
6 T^ 0, and if n is a positive integer, then
Proof. By applying the law for multiplying fractions, we have
(a\ n a a a ,, f , N a n
(b) =b'b ..... ^ (ton factors) =^-
For example,
/3xy^\ 2 _3^y^ _ 9o; 2 i/ 2fe
\ 2 2 / " 2 4 "" 2 4
An exponent affects only that quantity to which it is attached.
Thus, -5x(y*) 2 = -5xy Q , whereas (~5xy*) 2 = 25x 2 y*.
So far we have defined a n only when n is a positive integer. We
shall now introduce zero, negative integer* and rational powers in
such a way that they will obey the same laws which were proved
for positive integral exponents.
4-2. MEANING OF o
We shall define the zero exponent by the equation
(4-7) o = 1 (a 7* 0).
A few illustrations are :
= 5, (o - to)' a 1,
Sec. 4-2 The Laws of Exponents 89
If a in (4-2) is not zero and m = n, we get
a n
In this case, the quotient on the left equals 1, while the value of the
term on the right is a. Since a 1, by definition, the law of
division holds for n = m, as well as for m > n and n > m.
The student should note that (4-7) gives the only possible defini-
tion of a if the law expressed by (4-2) and (4-3) is to hold for the
zero exponent, as can be seen from the foregoing discussion.
We shall show that the definition a = 1 is consistent with the
five laws of exponents in Section 4-1 ; that is, we shall show that
these laws also hold when any exponent is zero. In the following
explanations, where a quantity occurs in a denominator, we assume
that it is not zero. Also, the exponents are assumed to be non-
negative integers.
Let us, for sake of discussion, suppose that n = in (4-1), that
' a m a n = a m ~*~ n .
Then we have . a . = ~ - . a o = a . . x =
OP a m+n = a m+0 = a m -_ a m . ^
Hence, the law of multiplication holds when n = 0. A similar
procedure will verify the law if m 0.
Now let us suppose that n = in (4-2), that is, in
1
= fr = = a m . and a m ~ n = a m ~ = a w .
a 1
Hence, (4-2) holds when n - 0.
If m = in (4-3) , we have
= =: , and = =
a n "" a n ~~ a n a n ~~ m ~~ a n ~ "~ a n
It is clear that in (4-2) m cannot be zero, and in (4-3) n cannot
be zero. Therefore, the law of division holds.
Suppose that n = in (4-4) , that is, in
(a m ) n = a mn .
Then
(a m ) n = (a m ) = 1, and a mn = a mt = a = 1.
If m = in (4-4) , we have
(a w ) n = (a) n = l n = 1, and a mn = a' n = a = 1.
Hence, the law for a power of a power holds.
90 The Laws of Exponents Sec. 4-2
Now consider (4-5), which is
(ab) n = a n b n .
If n = 0,
(a6) n = (a&) = 1, and a n b n = a6 = 1-1 = 1.
Finally, we let n = in (4-6), that is, in
/a\ n __ a^
W - 6 "
Then " t , a" a 1 t
1 > and = = = L
The demonstrations just given prove that the five laws of expo-
nents, originally stated for positive integral powers, are true for
all non-negative integral powers, and that the law of division is
true even when the exponents are equal.
4-3. NEGATIVE EXPONENTS
In order to extend the meaning of exponents to negative integers,
we define cr n by the following relationship :
(4-8) or* = 1 (a* 0),
where n is a positive integer.
Several illustrations are:
K-2 in 3 innn
5 ~ 5 2 ~ 25 10- 3 - 1U - 1UUU >
( a fa.)-2 =
v '
-
(a 6z) 2
V
As in Section 4-2, we shall show that our definition is consistent
with the five laws of exponents.
Let us first note that (4-8) is true even if n = or if n is a nega-
tive integer. If n = 0, then
-_()-!- I-!-!.
a _ a - 1 - j - a0 - fln
If n = p, where p is a positive integer, then cr n = a p = 1 /
a
n
cr p a
We shall use this result in the proofs that follow.
In order to extend (4-1), let m be a non-negative integer, and let
n be a negative integer, say, n = p. In this case it is also assumed
that a = 0. Then
1 a m
# m # n = d m cpp = a m =
a* a*
Sec. 4-3 The Laws of Exponents 91
If m S p, we have, by (4-2) for non-negative exponents,
/y W
a m a n = - = a m ~~ p = a m+(-p) _. a m+n^
a"
Ifm<p, we have, by (4-3) for non-negative exponents,
~~~ /tW
~~
A similar demonstration establishes (4-1) in case n ^ and
m < 0, or in case m < 0, and n < 0.
Proof of the extension of (4-2) rests on the validity of the law
of multiplication just established. If a =Q, and if m and n are
integers (positive, negative, or zero), then
a m 1
= a m = a m or n = a m ~ n .
a n a n
Although (4-3) is now an immediate consequence, it is not really
needed, in view of the general validity of (4-8). The demonstra-
tion just given allows the law of division to be stated as a single
relationship as follows :
fjTn
(4-9) ^ = a w ~ n (a 7* 0).
Thus, a single law applies, regardless of whether m > n, n > m, or
m = n, where m and n are arbitrary integers.
Now consider the law for a power of a power. In (4-4) let
n p, where p 0, while m ^ 0. Then
Also,
a mn a m-p -. a -mp = -
a mp
Hence, (4-4) holds in this case.
If m p, where p i^ 0, while n S 0, we have
(a m ) n (ar p Y ( J = and a mn = a ( ~ p)n = a~ pn =
Again (4-4) holds. If both m and n are negative, a similar proce-
dure is used, and the extension of (4-4) holds.
To extend the law for a power of a product, let n = p in (4-5),
where p ^ 0. Then
n- - _ * _ * nn--?-/)- 11 - 1
So (4-5) is verified for negative integral values of n,
To verify the exltended law for a power of a quotient, assume
that n = -p in (4-6), where p ^ 0. Then
(a\ n /a\~~ p 1 /a p b p , a n ar p b p
h) = (h) = P = */hP ~ ~P' anci fr ^ IT* = "5 "
92 The Laws of Exponents Sec. 4-3
Hence, (4-6) holds for negative integral values of n.
Thus, the laws of exponents hold for positive integral exponents,
zero exponents, and negative integral exponents. In Section 4-5
we shall consider the case of fractional exponents.
From the general validity of (4-8), it follows immediately that
a factor of the numerator or the denominator of a fraction can be
moved from the numerator to the denominator, or vice versa, pro-
vided only that we change the sign of its exponent. For example,
a 2 x 3 x^z 2
a 2 b~ 3 = TQ and o =
SCIENTIFIC NOTATION
We are now in a position to introduce certain simplifications
when operating with very large or very small numbers, as are
customarily used in scientific writing. Any positive number that is
greater than 10 or less than 1 may be written compactly by
expressing it in standard form, that is, by writing it as a number
that lies between 1 and 10 multiplied by a suitable positive or nega-
tive integral power of 10. Thus, 27,000 would be written 2.7 10 4 .
Similarly, 0.00031 would be 3.1 10~ 4 .
Example 4-1. The speed of light is 186,000 miles per second. Express this
number in scientific notation.
Solution: The given number 186,000 may be written as 1.86 10 5 .
Example 4-2. The rest mass of an electron is 9.11 10~ 28 grams. How many
zeros would he required between the decimal point and the first non-zero digit, 9, if
the number were written in decimal notation?
Solution: The exponent 28 means that we would have to move the decimal
point 28 places to the left from its present position. We would thus have to place
27 zeros to the left of the 9.
Example 4-3. If the sun is 9.3 10 7 miles from the earth, how long does it take
light to reach the earth from the sun?
Solution: As given in Example 4-1, the speed of light is 1.86* 10 5 miles per
9 3 . iQ7
second. Therefore, the required time is i "b fi . i?^ = 5 10 2 = 500 seconds =
8 minutes 20 seconds.
4-5. RATIONAL EXPONENTS
We shall now extend the meaning of exponents from integers to
rational numbers. Here again we shall make the extension in such
Sec. 45 The Laws of Exponents 93
a way that the laws for positive integral exponents will be
preserved.
Suppose that a is a real number and that n is a positive integer.
Let us assume that a 1/n has meaning and that (4-4) applies. Then
it would be true that
(4-10) (a 1 /")" = a' 1 /^ = a 1 = a.
This says that the nth power of a l/n would have to be a, or in other
words that a l/n would be what is called an nth root of a. For
example, (4-10) would yield
(a 1 / 2 ) 2 = a, and (a 1 / 3 ) 3 = a.
Real nth Roots of a. Before defining a l/n , let us examine the sit-
uation with respect to the existence of nth roots of a given number
a. The following results may be proved with the help of the theory
of equations.
Case I. If n is an even integer and a is a positive real number,
there are two real numbers that satisfy the equation r n = a. One
of these is the positive nth root of a, which is denoted by tya. The
other is the negative nth root of a, which is denoted by ^/a. We
may also denote these two numbers together by ^/a.
Case II. If n is an even integer and a is a negative real number,
no real nth roots exist, since no even power of a real number can
be negative.
Case III. If n is an odd integer and a is a positive real number,
there is one real (positive) value of r such that r n = a. In other
words, if n is odd, there is a real positive nth root, which is denoted
by^a.
Case IV. If n is an odd integer and a is a negative real number,
there exists one real (negative) value of r such that r n a. That is,
if n is odd, there is a real negative nth root, which is denoted by
y*.
Case V. If n is any positive integer and a is zero, there is only
one real nth root, and this root is zero.
Thus, the definition of an nth root of a is valid under all condi-
tions except when a is negative and n is even. In this situation, no
real nth roots exist. (However, the introduction of complex num-
bers in Chapter 11 will allow us to eliminate this exception.) We are
now ready for the following definition.
Definition. If a is a non-negative real number and n is a positive
integer, a lfn designates the non-negative nth root of a, or ^/a. If a
is negative and n is an odd positive integer, then a 1/n designates the
real nth root of a, or ^/cT.
When a is negative and n is even, a 1/w is undefined:
94 The Laws of Exponents Sec. 4-5
Meaning of a m/n . Let a be a given real number, n a positive
integer, and m an integer. If a m/n has meaning, and if (4-4) holds
for fractional powers, then a m/n = a (1/n)TO = (a 1/n ) m . Under these
assumptions, then, a m/n would be the mth power of a 1/n . It is
natural to state the following definition.
Definition. If n is a positive integer, if m is any integer such that
the fraction m/n is in lowest terms, and if a is a real number which
is assumed to be non-negative when n is even, then a m/n designates
the mth power of a 1/n , that is, the mth power of ^/a. Hence,
(4-11) a mln = (a 1/n ) m .
If the fraction m/n is not in its lowest terms, it is first reduced to
lowest terms, and (4-11) is then applied.
When a is given, the value of a m/n depends only on the value of
the fractional exponent, not on the particular values of m and n.
Thus,
2 4/2 = 2 2 = 4, 2 6 ' 8 = 2 3 ' 4 , and (- 2) 2 / 6 = (- 2) 1 / 3 = ^^2.
In the last example it would be incorrect to apply (4-11) directly,
since ( 2) 1/6 has no meaning.
It may be shown that, if a is positive,
(4-12) a mfn = <\/a.
The proof is omitted.
We shall also omit the details of the procedure for showing that
the five laws of exponents hold for rational exponents and non-
negative bases. The reader is cautioned against using the laws for
negative bases, since some fail under certain conditions.
To summarize the results now established, we restate the laws of
exponents here for easy reference. It is assumed that a and b are
non-negative real numbers, and that m and n are rational numbers.
Furthermore, if either a or 6 appears in a denominator or raised
to a negative or zero power, it is assumed to be different from zero.
Law of multiplication: a m a n = a m+n .
a m
Law of division: = a m ~ n .
a n
Law for a power of a power: (a m ) n = a wn .
Law for a power of a product: (ab) n = a n b n .
(a\ n a n
-J 5= 7-
Law for reciprocal: ar n =
a n
Zero power: a = 1.
Sec. 4-5 The tows of Exponents 95
Note that the radical notation can be replaced by the simpler
and much more convenient exponential form. Everything that can
be done with the radical notation in the simplification of roots of
numbers and in operations involving roots can be done much more
naturally by means of the exponential notation. A few illustrations
of the meanings and uses of exponential forms follow :
x 1 ' 2 = Vx, if * ^ 0; (- 27) 1/3 = \/^~W = - 3;
- (32) 1 / 5 = - X/32 = - 2; - -tf/5 5 = - a 2 ;
32 2 / 5 = (\X32) 2 = 2 2 = 4; -
(16s 4 ) 1 ' 2 yi6x* 4x 2
The following examples illustrate the applications of the laws of
exponents to the solution of problems involving radicals.
Example 4-4. Compute the value of \/2 \/2, and write the result in expo-
nential form.
Solution: Using exponential notation and the laws of exponents, we have
2i/3 . 21/4 = 2 4 /* 2 3/12 = 2 4/12+3/12 = 2 7/12 .
Example 4-5. Remove all possible factors from the radical $
Solution: We may proceed as follows:
= (2 3 4 z 4 2/ 2 ) 1/3 = 2 1/3 3 4/3
= 3x 2 1/3 3 1/3 x 1/3 2/ 2/3 = 3
Example 4-6. Use the laws of exponents to express ^/x ^/y by using only
one radical.
Solution: Changing to fractional exponents, we have
Example 4-7. Rationalize the denominator in the fraction -J7=f
Solution: To write an equivalent fraction in which no radical appears in the
denominator, we proceed as follows: __
Example 4-8. Rationalize the denominator of the fraction -=
5 v 3
Solution: We use the relationship (a + b) (a b) = a 2 b 2 to remove the
radical from the denominator. Thus,
1 = 1 5 + \/3 = 5 + V3 = 5 + V3
5 - V3 " 5,- V3 ' 5 + V3 " 25 - 3 22
96 The Laws of Exponenfs Sec. 4-5
. ^2
/a? - x 2 + -
Example 4-9. Change 2 __ ^ to a simple fraction.
Solution: Write the expression in exponential form, as follows:
a' 2 - x 2
Then, multiplying the main numerator and the main denominator by (a 2 - x 2 ) 112 ,
have ^ _ <r2 + ^ 2
(a 2 - z 2 ) 3 / 2 (a 2 - .r 2 ) 3 ' 2
Example 4-10. Express (x 2 + a 2 ) 3 / 2 + Zx 2 (x 2 + a 2 ) 1 ' 2 in a factored form,
Solution: Rewrite the expression as
(x 2 + a 2 ) 112 (x 2 + a 2 ) + Zx 2 (x 2 + a 2 ) 112 .
Removing the common factor (x 2 + a 2 ) 1 / 2 , we obtain
(x 2 + a 2 ) 112 [x 2 + a 2 -j- 3^ 2 ] = (x 2 + a 2 ) 1 / 2 (4z 2 + a 2 ).
EXERCISE 4-1
In each of the problems from 1 to 20, perform the indicated operations and
eliminate all zero and negative exponents.
1. 3x*y. 2. ari". 3. (jV 3 - 4. 10- 2 .
5- Trb' 6- UJ* 7. (9)". 8.
9.
13. (a;V)*(a:-V 4 ) 2 . 14. (x 1 '^- 8 ) 4 ^- 1 ' 8 ) . 15. (x*' 2 -hi/ 1/2 ) 2 .
16. x-i 4- y-1. 17. (x + I/)- 1 - 18. (a; + 2/)- 1/3 (x + j/)i/.
10 Lz^l!!!. 20 ;
-
Write each of the following expressions in exponential form. Remove all possible
factors from the radical and, wherever necessary, rationalize the denominator.
21. V80.
s- V* 2 -
Sec. 4-7 The Laws of Exponents 97
-/I T 2 I ^
-v/T"
^2
A_t\ \./ t } / r T 9 1 * t/ ^"*" * v 'i .
46 V
\/2x - x 2
1 - X 2
18. M 4- 1V3/2 4- r2^
1 4. -y^l -f- X 2
47. - V 1 + ?f .
1)1/2.
4-^4. 2)2
49. (2 - x 2 ) 5 / 2 + s 2 (2 - z 2 ) 3 / 2 . 50. Or 2 - 3) I/2 - x*(x 2 - 3)- 1 ' 2 .
4-6. THE FACTORIAL SYMBOL
The product of all positive integers from 1 to n inclusive is called
"n factorial" or "factorial n" and is represented by either of the
symbols n ! or /.n. Thus, if n is a positive integer,
n\ = 1 -2-3 ..... (n - l)-n.
For example,
3! = 1 2 - 3 = 6; 5! = 1 2 3 4 5 = 120;
6! = 5!. 6; j = 71; r! = [(r - l)!]r.
4-7. THE BINOMIAL THEOREM
The statement known as the binomial theorem enables us to
express any power of a binomial as a sum of terms without per-
forming the multiplications.
By actually performing the indicated multiplications, we find that
(a + 6) 2 = a 2 + 2ab + 6 2 ,
(a + b) 3 = a 3 + 3a 2 fe + 3ab 2 + 6 3 ,
(a + 6) 4 = a 4 + 4a 3 6 + 6a 2 6 2 + 4a& 3 + 6 4 .
These formulas may be rewritten in the following manner, so as
to suggest a general rule 1 :
(a + &) 2 = a 2 + ?a6 + ~fc 2 ,
(a + 6) 3 = a 3 +ja 2 6 + j^|a6 2 + |ff^& 3 ,
(a + *)4 = a 4 + + if
Applying this suggested rule to (a + 6) 5 , we obtain
(a + 6)5 = a* + a 4 6 + a 3 6 2 + f
S-4-8-2 4 5>4>3>2>1
^ 1 2 3-4 a ^1.2-3-4-5
1 The justification for writing the expressions on the right in this form will
be found in Chapter 17, where the binomial coefficients are "given in terms of
the combination formulas.
98 The Laws of Exponents Sec. 47
Upon simplification of coefficients, we get (a + b) 5 = a 5 + 5a 4 b +
10a 3 b 2 + 10a 2 6 3 -I- 5ab* + ft 3 , which is the same result as that
obtained by multiplying (a + &) 4 by (a + 6) .
Each of the expressions on the left is of the form (a + b) n , in
which the exponents 2, 3, and 4 of (a + 6) are special values of n.
If we let n denote the exponent of (a + 6) in each of the expres-
sions on the left, we note that the expansion of (a 4- b) n contains
n + 1 terms with the following properties :
1. In any term the sum of the exponents of a and b is n. Also,
the first term is a n and the last term is b n .
2. The exponent of a decreases by 1, and the exponent of 6
increases by 1, from term to term.
3. The denominator of the coefficient in each term is the fac-
torial of the exponent of b in that term.
4. The numerator of the coefficient in each term has the same
number of factors as the denominator. Specifically, wherever 1
appears in the denominator, write n directly above it in the
numerator ; wherever 2 appears in the denominator, write n 1
directly above it in the numerator; and so on. Thus, in (a + 6) 5 ,
the number above 1 is 5, and the number above 2 is 4.
Assuming that these properties hold for all positive integral
values of n, we have
(4-13) (a + b) n = a* + ^ a^b + n(n "^ a - 2 b 2
i i z
1 * ^ * O
This result is the binomial formula. So far we have verified this
formula only for n = 2, 3, 4, and 5. In Chapter 16, we shall prove
the binomial theorem, which states that the formula is true for all
positive integral values of n.
The following example shows the procedure for the expansion of
Example 4-11. Expand (x - 2?/) 6 by the binomial theorem.
Solution: The required expansion will be obtained by letting a = x, b = 2y,
and n = 6. We begin by setting up the following pattern of n + 1 terms :
x 6 + - ( - 2y) + - x*( - 2y) 2 + - x*( - 2y)
+ - x*( - 2t/) 4 + - x( - 2y) + ( - 2).
Sec. 4-8 The Laws of Exponents 99
The exponents of b in the second, third, fourth, and fifth terms are 1, 2, 3, 4, and 5,
respectively. Hence, remembering that the last term is 6 n , we may fill in the
numerators and denominators of the coefficients as follows:
By simplifying, we obtain the following result:
(x - 22/) 6 = z 6
4-8. GENERAL TERM IN THE BINOMIAL EXPANSION
If we wish to write any particular term of the expansion of
(a + b) n without considering any of the other terms, a study of the
binomial formula in (4-13) Section 4-7 will reveal the' following
facts :
In every term the exponent of 6 is one less than the number of
the term. Thus, in the (r-f l)th term, the exponent of 6 is r.
(The expression for a particular term is simplified slightly if the
number of that term is called r + 1, rather than r.)
The sum of the exponents of a and 6 is n in each term. For
the (r + l)th term the exponent of a is n r.
The denominator of the coefficient in the (r + l)th term is rl,
since it is the factorial of the exponent of b.
The numerator of the coefficient has the same number of fac-
tors as the denominator. In the (r + l)th term, it is the product
n(n 1) (n 2) (n r + l).
We obtain, then, for the (r + l)th term of the expansion
n(n - 1) (n - 2) (n - r + 1) alMftr
r!
Example 4-12. Find he sixth term of (3x - y 2 ) 8 .
Solution: Here a = 3z, 6 = - y 2 , and n = 8. Since r + 1 = 6, the exponent of
6 is r = 5. Hence, the exponent of a is n 5 = 3. Therefore, the sixth term is
8<7 ' 6 ' 5 ' 4 oxr( ) =
z Q . . \ox) v if ;
1 00 The laws of Exponents Sec. 4-8
EXERCISE 4-2
In each of the problems from 1 to 16, reduce the given fraction to lowest terms.
9.
(91) (31).
(3-4)!
10! .
3 (11) (30.
. 3! +4!
4 (7!) (8!)
(5!) (6!)
a n! -
- (3!) (7!)
3(4!)
n!
6 * (3!) (4!)
(3!) (4!)
a< (n - 1)1
(n-2)!
13.
15.
17. Sh/w
< + l)!(!!l!
1L (n - 1)1
. [(n + 1)!] 2
! 31 (n- 3)!
!
(n!) 2
n!(n + l)l
n!(n -2)!
fi n!(n -2)!
( \~ l^n ^1)1 - 2)
[(n -1)!]*
In each of the problems from 18 to 32, expand the given expression by the
binomial theorem. (Hint: In problems 29 thru 32, first consider the first two terms
in parentheses as a single quantity).
18, (x + y)*. 19. (x - I) 7 . 20. (a - 26). 21. (2a* - 362)3.
(7/2 /ll/3\6
^r+V ) 2. (*+
iC 2r '
30. ( + 2y + ) 2 . 31. (x* +x + I)*. 32. (o - a - I) 3 .
In each of the problems from 33 to 42, find the indicated term.
33. (1 - s) 8 , 8th term. 34. (m + n)", 10th term.
35. (a + 26) 12 , 5th term. 36. (a - 6), 3rd term.
(x 2\ 8
5 ) > 4th term. 38. (x - 2/) l , term involving y 4 .
t X/
(x v*\ n 1
-- ~ ] , term involving
y * y
41. (V^ - vV) 12 , middle term. 42. (2x - ?/) 7 , middle terms.
5
Logarithms
5-1. DEFINITION OF A LOGARITHM
We shall assume here that the laws of exponents stated in
Chapter 4 for rational exponents ark valid also for irrational
exponents. The definition of a base raised to an irrational power is
beyond the scope of this book. However, let us make the assump-
tion that, if b and x are real numbers, with b positive, a corre-
sponding number designated by" b 33 exists. Without giving an
explicit rule for computing b x , let us assume that all laws of expo-
nents established in Chapter ~4 are valid generally for real powers.
Finally, let us assume that, corresponding to any two positive real
numbers & and n, where b = 1, there exists a unique real number x,
such that n = b x . We can then give the following definition.
Definition. If n = b x , where 6 is a positive real number different
from 1, then x is called the logarithm of n to the base b. We write
x = log & n. The following table shows both forms of several equiva-
lent statements.
Exponential
Form
2 3 =8 .
41/2 :
= 2
1
3 = 8l
5
= 1
Logarithmic
Form
Iog 2 8 = 3
Iog 4 2
1
" 2
"*4=-*
logs
1 =0
We shall restrict n to positive numbers, since negative numbers
do not have real logarithms. '
For any positive base &, we have 6 = 1 and 6 1 = 6. Hence, it fol-
lows from the definition of a logarithm that
log b 1 = and log& b = f l.
Another valuable, relationship results from combining the two
equations n = b* and x = log & n. Replacing x in the first equation by
its value from the second equation, we have
For example, 2 10 ** 8 = 8, and 10 10 *i<> * = x.
101
1 02 Logarithms Sec. 5-1
Example 5-1. Find n, if logs n = 2.
, Solution: Write the given equation in exponential form, as follows:
n = 3 2 .
Hence, n = 9.
Example 5-2. Find the base 6, if log* 4 = 2/3.
Solution: In exponential form, the given equation is 6 2/3 = 4. Raise both sides
to the 3/2 power and recall that 6 > 0. Then
(52/3)3/2 = & = 43/2,
Therefore, 6=8.
Example 5-3. Find x, if logi/ 8 32 = x.
Solution: Writing the equation in exponential form, we have
Express 1/8 and 32 as powers of 2, and get 1/8 = 1/2 3 = 2~ 3 , and 32 = 2 5 . Hence,
(2-3)* = 2 5 , or 2~ 3J! = 2 5 .
From this, we have - 3x = 5, and x = - 5/3. Therefore, logus 32 = 5/3.
5-2. LAWS OF LOGARITHMS
Since a logarithm is an exponent with respect to a given base,
the rules for operating with logarithms are the same as the laws
of exponents. These laws, expressed in terms of logarithms, have
the following form.
Law I. The logarithm of a product equals the sum of the loga-
rithms of its factors. The logarithmic form is
(5-1) log& (m n) = log& m + Iog6 n.
Proof: To prove this equation, let
x = logb m and y = log& n.
Then
m = b x and n = b v .
Multiplying, we have
mn =b*+v.
Hence, .
(mn) = x + y = log& m + log& n.
Law II. The logarithm of a quotient equals the logarithm of the
dividend minus the logarithm of the divisor. The logarithmic
form is
(6-2) Iog6 = Iog 6 m - Iog 6 n.
Sec. 5-2 Logarithms 103
Proof: The proof follows : Let
x = logb m and y = Iog6 w.
Then ,, , ,
m = o* and ?i = o^.
Dividing, we have _
= b<-.
n
Hence, , m ^
iog& ( ) = x - y = log b m log b n.
\ 71 /
Law III. The logarithm of a power of a number equals the expo-
nent times the logarithm of the number ; that is,
(5-3) logo (n*) = * log b n.
Proof: The first step in the proof is to let
x = logb n.
Then .
n = b*.
Raise both sides to the kth power and obtain
n k = (b x ) k = b kx .
This relationship, when written in logarithmic form, becomes
Iog6 (n k ) = kx.
Replacing x by its value, we have
log& (n k ) = k Iog6 n.
The student should note carefully the difference between log& (n k )
and (log&n) fc .
Law IV. The logarithm of a root of a number equals the loga-
rithm of the number divided by the index of the root ; that is,
(5-4) Iog 6 \fn = ^ log b n.
Proof: This equation follows as a corollary of law III. By the
definition of a fractional exponent, we have <tyn = n l/k . Hence, by
law III,
Iog 6 tyn = Iog6 (n 1/fc ) = Iog6 n.
\ A/51
Example 5-4. Express loga -~^- as a linear combination of logarithms.
Solution: Iog 2 = Iog 2 VSl - Iog 2 3 4 = Iog 2 (3 17V' - Iog 2 3*
= Iog 2 S 1 ^ + lo g2 17 1 / 2 - Iog2 3 4 .
104 Logarithms Sec. 5-2
Example 5-5. Express 2 logio 3 - ~ logio $ + logio y as a single logarithm.
Solution: 2 logio 3 - = logio x + logio y = logio 3 2 - logio a: 1 ' 2 + logio 2/
J = logio (32 y) - logio x"*
Example 5-6. Transform the equation logo x + y = Iog sin x into an equation
free of logarithms.
Solution: By transposing, we get
, , , sin x
y = loga sin * - logo x = logo ---
Change to the following exponential form :
sin x
EXERCISE 5-1
In each of the problems from 1 to 12, write the equation in logarithmic form.
1. 2 3 = 8. 2. 2 = 64. 3. 3 4 = 81. 4. 10 = 1.
5. 10 3 = 1000. 6. 10- 3 = 0.001. 7. 256 1/8 = 2. 8. 216 l/3 = 6.
9. 100- 6 = 10. 10. y = e*. 11. 10" = x. 12. 10 10 * * = x.
In each of the problems from 13 to 21, write the equation in exponential form.
13. logs 64 = 2. 14. logs 125 = 3. 15. Iog 2 ^ = - 6.
16. log* ggg = - 4. 17. Iog 7 343 = 3. 18. Iog 9 729 = 3.
19. logio 10,000 = 4. 20. logio 0.0001 = - 4. 21. Iog 4 8 = 3/2.
In each of the problems from 22 to 33, find the indicated value of x.
22. Iog 9 3 = Z. 23. Iog 2 64 = x. 24. Iog 4 x = 0.
25. log* 4=2. 26. logo.s x = - 1. 27. Iog 3 x = 1.
28. log* 81 = 4. 29. log, 100 = - 2. 30. log, ~ = 5.
7
31. logo 243 = x. 32. Iog 64 x = - ^ 33. log* x = 2.
In each of the problems from 34 to 39, use the laws of logarithms to write the
expression as a single logarithm.
34. logfr 2-3 log& 5 + log& 7. 35. log& 4 -f log& TT - log& 3+3 log& r.
191 2^
36. \ log, 7 + 1 Iog 4 4 + 1 log* 3. 37. - 5 log 23 + 12 log* =f
ODD &
38. 3 logs 2 + logb 13-2 log& 5.
Sec. 5-4 Logarithms 105
39. logb (u - Vu 2 - a 2 ) - - logb (w + Vu 2 - a 2 ) + log& a.
z &
40. Find the logarithm to the base b of the area of a circle in terms of the logarithms
of TT and the radius.
41. The time T for a pendulum of length / to make one oscillation is T = IT \/ - >
where g is a constant representing the acceleration due to gravity, a) Find
log?, T in terms of the logarithms of IT, I, and g. b) Find log& I in terms of the
logarithms of w, T, and g.
42. The area of a triangle with sides of length a, 6, and c is given by the formula
K = Vs( s a) (s b) (s c), where s is the semi-perimeter ^ (a +6 + c).
Find Iog6 K in terms of the logarithms of combinations of a, b, and r.
43. The positive geometric mean G of n positive numbers x\,x*, . x n is defined
by the relationship
, n
log* G =
Show that G = \/XiX2 - - x n .
5-3. SYSTEMS OF LOGARITHMS
As we mentioned in Section 5-1, any positive number 6 different
from 1 may be used as a base in a system of logarithms. However,
only two bases are widely used in practice.
The common, or Briggs, system of logarithms, named for Henry
Briggs (1556-1631), employs the base 10 and is used for ordinary
computations.
The natural, or Napierian, system of logarithms, named for
John Napier (1550-1617), is generally used in calculus and theo-
retical work, and employs the more convenient irrational base
e = 2.71828
In this book, when the base is not indicated, it is understood to
be 10. Thus, log n means logic n, and the word logarithm will mean
common logarithm unless otherwise stated.
5-4. COMMON LOGARITHMS
In Table 5-1, we begin with a list of powers of 10, give equivalent
logarithmic forms, and from these determine the form of the loga-
rithm of a number that is not an exact power of 10. It should be
mentioned that the logarithm is an increasing function; that is,
as n increases, log n increases. Another way of stating the condi-
tions is to say that if a > b then log a > log 6.
106
Logarithms
TABLE 5-1
See. 5-4
Exponential form
Logarithmic form
Logarithm
of the number
10 3 = 1000
log 1000 = 3.000
"-log 354.
= 2 + decimal
10 2 = 100
log 100 = 2.000
<-log 35.4
= 1 + decimal
10 1 = 10
log 10 = 1.000
<-log3.54
= + decimal
10 = 1
log 1 = 0.000
<-log 0.354
= 1 + decimal
10- 1 = 0.1
log 0.1 = - 1.000
<-log 0.0354
= 2 + decimal
lo- 2 = o.oi
log 0.01 = - 2.000
-log 0.00354
= 3 + decimal
io- 3 = o.ooi
log 0.001 = - 3.000
From Table 5-1 it can be seen that the following statements are
true :
The logarithm of an integral power of 10 is an integer.
The logarithm of a number which is not an integral power of 10
consists of two terms or parts : an integral part, called the charac-
teristic; and a positive or zero decimal part, called the mantissa,
which is determined from a table of mantissas.
Thus, since log 10 = 1 and log 100 = 2, we may expect the loga-
rithm of any number between 10 and 100, that is, a number between
IO 1 and IO 2 , to be 1 plus a positive decimal part. For example, we
shall find that the logarithm of 35.4, which number lies between 10
and 100, is equal to 1.5490, to four decimal places. In this case, the
characteristic is 1 and the mantissa is .5490.
5-5. RULES FOR CHARACTERISTIC AND MANTISSA
A study of Table 5-1 reveals that the characteristic changes as
the position of the decimal point changes in the sequence of digits
0035400. The first entry in the column headed "Logarithm of the
number" is
log 354 = 2. + decimal.
Sec. 5-5 Logarithms 1 07
In the number 354, or 354.0, the decimal point is two places to the
right of the first non-zero digit, 3 (reading from left to right) ;
the corresponding characteristic is 2.
The second entry is
log 35.4 = 1. + decimal.
In this number, 35.4, the decimal point is one place to the right of
the first non-zero digit (reading from left to right) ; the corre-
sponding characteristic is 1.
Similarly, we note that the zero characteristic corresponds to the
position of the decimal point immediately following the first non-
zero digit. This position of the decimal point is called the standard
position. We may now formulate the following rule for
characteristics :
Rule for Characteristics. If the decimal point is in standard posi-
tion, the characteristic is zero. For every other position of the
decimal point, the characteristic is equal to the number of places
the decimal point has been shifted from the standard position. The
characteristic is positive if the shift is to the right, and is nega-
tive if the shift is to the left.
We shall now see that the mantissa remains the same for all
numbers having the same sequence of digits. Let us again consider
the sequence of digits 0035400. Any number containing this
sequence can be written 3.54 10 n , where n is a positive or negative
integer or zero and depends on the position of the decimal point.
Suppose that we consider the form log 3.54 = 0.5490. Then the
logarithm of any number containing this sequence is
log (3.54 10 n ) = log 3.54 + log 10 n
= n + log 3.54
= n + 0.5490.
Thus, a shift of the decimal place in the number affects only the
characteristic n, and the mantissa remains the same for the same
sequence of digits.
EXERCISE 5-2
In each of the problems from 1 to 16, find the characteristic of the logarithm of**
the given number.
1. 34.63. 2. f 3.463. 3. 34630. 4. 268.1.
5. 0.1340. 6. 2637. 7. 0.00346. 8. tan 428'.
7 821
9. sin 6341'. 10. 0.000001. 11. 378364. 12. ~~'
lUjUUU
13, cot 8113'. 14. sin 8453'. 15. cos 6143'. ' 16. sec 24 b 8'.
1 08 Logarithms Sec. 5-5
In each of the problems from 17 to 24, place the decimal point in the sequence of
digits 7314 corresponding to the given characteristic.
17. 3. 18. - 2. 19. 0. 20. 1.
21. 6. 22. - 5. 23. - 3. 24-1.
5-6. HOW TO WRITE LOGARITHMS
As stated in Section 5-4, the mantissa of a logarithm is always
positive or zero, whereas the characteristic may be a positive or
negative integer or zero. A positive characteristic or a zero char-
acteristic can readily be combined with a given mantissa. For
example, the logarithm of 354 is written 2.5490. But when the
characteristic is negative, say fc, where 1 ^ k ^ 10, it is more con-
venient to write it in the form (10 k) 10. Let us consider the
logarithm of 0.00354. The characteristic is 3, but the mantissa is
regarded as positive. We could write log 0.00354 = 34- 0.5490.
For convenience in computation, however, we write log 0.00354 in
the form (10 - 3) + 0.5490 - 10 = 7.5490 - 10, or 17.5490 - 20,
and so on.
Note. It would be incorrect to write log 0.00354 = 3.5490, for
this notation means 3 0.5490 and would imply that the mantissa
is negative. To perform certain computations, it is convenient to
write the logarithm 7.5490 - 10 in the form -2.4510, which equals
2 0.4510. It is important to note that the decimal part of the
number 2.4510 is not the mantissa of the logarithm of 0.00354,
since it is not positive.
5-7. HOW TO USE A TABLE OF MANTISSAS
The following examples will illustrate the procedure in finding
the logarithm of a number with the aid of a table of mantissas. The
student should work through each example, determining the char-
acteristic from the position of the decimal point in the number and
determining the mantissa by referring to Table III at the end of
this book.
Example 5-7. Find log 46.7.
Solution: The characteristic is -f 1. To find the mantissa, locate 46 in the
column in the table headed N, and then go to the right to the column headed 7.
Here we find the mantissa .6693. So the complete result is log 46.7 = 1.6693.
Interpolation. If the number consists of more than three digits,
the mantissa is found from Table III by means of interpolation.
Since the method of interpolation is the same as that described in
Section 3-10 for the table of trigonometric functions, there will be
no further discussion of it here.
Sec. 5-7
Logarithms
109
Example 5-8. Find log 0.03426.
Solution: The characteristic is 2. The mantissa is found by interpolation,
since the number 3426 has more than three digits. It lies of the way between
the mantissas of 3420 and 3430, as shown in the accompanying tabulation:
Number Mantissa
10
f 3420
.5340 1
6
> r
1 3426
.5340 + x j
3430
.5353
13
Since the difference between the mantissas of the two numbers in the table is 13,
we have A
x =
This is rounded off to 8, and the amount to be added to 0.5340 is given by x =8.
Hence, the mantissa is .5348 and log 0.03426 = 8.5348 - 10.
Finding Antilogarithms. The number which corresponds to a
given logarithm is called the antilogarithm. That is, if log n = x,
then n is the antilogarithm of x and is written antilog x.
Example 5-9. Find n, if log n = 1.8710.
Solution: Search through the body of Table III to locate the mantissa .8710.
The corresponding number, from the columns headed N and 3, is 743. Since the
characteristic is 1, n = 74.3.
Example 5-10. Find antilog 7.5349-10.
Solution: The mantissa .5349 is not in Table III but lies between .5340 and
.5353. To these correspond, respectively, numbers whose digits are 3420 and 3430.
We may indicate the work in tabular form as follows:
Number Mantissa
3420 .5340
3420 + x
10
.5349
9
13
3430 .5353
From this, we see that *
. -JL
10 ~ 13 '
Therefore, x = 6.9, or 7 after rounding off. Hence, the sequence of digits in the
desired number is 3427. Since the characteristic is - 3, the untilogarithm of
7.5349-10 is 0.003427.
110
Logarithms
EXERCISE 5-3
Sec. 5-7
In each of the problems from 1 to 30, find the common logarithm of the given
number.
1. 35.
2. 98.
3. 105.
4. 0.0843.
5. 0.00621.
6. 9.63.
7. 23,100.
8. 12.3.
9. 0.354.
10. 0.0781.
11. 0.0663.
12. 1,630.
13. log 7.03.
14. log 95.5.
15. log 695.
16. log 6.31.
17. 0.007001.
18. 0.003821.
19. 0.7777.
20. 7,437.
21. 3.142.
22. 1.414.
23. 0.08788.
24. 15.46.
25. cos 1613'.
26. sin 1018'. 27. tan 4133'. 28. sec 6416'. 29. cos 8214'. 30. cob 3116'.
In each of the problems from 31 to 50, find the antilogarithm of the given number.
31. 1.6665. 32. 4.4857. 33. 9.4183-10.
35. 2.7024. 36. 7.7388-10. 37. 9.4409-20.
39. 1.8401. 40. 3.9552-10. 41. 2.4658.
43. 9.7367. 44. 4.9960-10. 45. 8.7863-10.
47. 0.6584. 48. 3.0150. 49. 5.0300-10.
Solve for x in each of the following equations:
51. 10* = 4. 52. 10" = 2.019.
54. 10 - 1 = x. 55. ^/10 = x.
57. ^10*= a;. 58. 10 1 - 314 =x.
60. 10-*' 2 = 0.0123. 61. 10 1 -* = 0.2346.
34. 0.0645.
38. 6.3404-10.
42. 1.9501.
46. 9.8821-20.
50. 0.1504.
53. 10 2 * = 7.132.
56. 10^ = x.
59. 10-* = 0.003146.
62. 10 2 *- 3 = 0.6735.
5-8. LOGARITHMIC COMPUTATION
The fundamental laws of logarithms given in Section 5-2 are
applied in the following examples to illustrate the application of
logarithms to computation.
Example 5-11. Find the product (0.0246) (1360).
Solution: Let x = (0.0246) (1360). Then
log x = log 0.0246 + log 1360.
log 0.0246= 8.3909-10
log 1360 = 3.1335
logo; = 11.5244-10
= 1.5244.
Hence, by interpolation, we have x = 33.45.
Sec. 5-8 Logarithms 1 1 1
Example 5-12. Evaluate (0.506)- 1 ' 3 .
Solution: Let x = (0.506) -' = / 05( L 1/3 . Then
logs =logl - log (0.506) 1 ' 3
= log 1 - (1/3) log 0.506
= log 1 - (1/3) (29.7042-30)
= log 1 - (9.9014-10).
log 1 = 10.0000-1Q
1/3 log 0.506 = 9.9Q14-1Q
log x = 0.0986.
Therefore, x = 1.255 by interpolation.
Alternate Solution: Let Z = (0.506) - 1 / 3 . Then
log x s= - (1/3) log 0.506
= - (1/3) (29.7042-30)
= - (9.9014-10)
= - ( - 0.0986)
= 0.0986.
Therefore, x = 1.255 by interpolation.
n i r 10 ^ i * (0-352) (1.74)2
Example 5-13. Evaluate =P
-^0.00526
Solution: Let x denote the desired value. Then
log x = log 0.352 + 2 log 1.74 - (1/3) log 0.00526.
We find that log 0.352 = 9.5465-10, log 1.74 = 0.2405, and log 0.00526 = 7.7210-10.
log 0.352 = 9.5465-10 (1/3) log 0.00526 = (1/3) (27.7210-30)
2 log 1.74 = 0.4810 = 9.2403-10.
log numerator = 10.0275-10
log numerator = 10.0275-10
log denominator = 9.2403-10
log x = 0.7872.
Interpolating, we have x = 6.126.
Example 5-14. Evaluate
Solution: Let x = ( 7=7 ) . Then
\174/
log x = 1.14 [log 253 - log 174].
log 253 = 2.4031
log 174 = 2.2405
Then - 1626 '
log x = 1.14 (0.1626) = 0.1854.
Therefore, x = 1.532.
112
Logarithms
Sec. 5-8
EXERCISE 5-4
In each of the problems from 1 to 30, perform the indicated computation using
logarithms.
1. (3.142)(2.718).
29.34
d * 68^5*
7. \/(0.003468) fi .
9.
o J/
r
(8,321,000)
(36,250) 4
11. (63.84)2(0.0134).
13. V(168.3) (14.21).
15. V(23,310) 2 - (20,180)2.
(Hint: Factor the radicand.)
19.
567.87
1 3 * 5 7 19 31
2 4 8' 16 32* 64'
21 A/ < 8L68 > 4
V (8.013) (0.0
(8.013) (0.034)
23. (1
25. v
27.
- (0.8123)- 3 ' 4 .
0.08614.
29, [(3.864)-3.i3 +
2. (13.25) (26.80).
4. (0.8134) 1/3 .
6. (16.83) 3 '*.
ft 836 - 1
42,860 "
10. (4.313)(3,068)(0.000642).
12. (8.364)(321.5) -K- 42.63).
14. V(213.6) 2 (43.98) 2 .
is 3,642^ v / (21.36) 3
*O "/i no>
(1,083) 4 (0.0813) 3
20 1 4/3188" f
38.63 K 0.8103
22. (1.08) 10 .
24. 3,648 (1.03) 3 6.
26. 0.083 ' 412 .
28. log 16.84 - ^483.6.
30.
(83.14)-* - (0.8134) 2 '*
31. If e = 2.718, find log e, log <\A, log - i e', and IT*.
c
32. Find the geometric mean of 564.3, 8634, 0.1349, 8.316, and 42.61. (Hint: See
Problem 43, Exercise 5-1.)
33. Find the area of a circle of radius 6,381 feet.
34. The volume of a sphere is V = 5 irr*. Find the volume of a sphere of radius
3621 feet.
35. Find the radius of a sphere whose volume is 8423 cubic feet.
36. Find the length of a pendulum which makes one oscillation in 1 second, if
g = 980 centimeters/sec 2 . (Hint: See Problem 41, Exercise 5-1.)
Sec. 5-9 Logarithms 113
37. Find the area of a triangle with sides 6,384 feet, 5,680 feet, and 2,164 feet long.
(Hint: See Problem 42, Exercise 5-1.)
38. The stretch s of a wire of length I and radius r by a weight m is given by the
relationship s = |r > where g is the gravitational constant and k (Young's
modulus) is a constant for a given material. Find how much a copper wire of
length 120 centimeters and of radius 0.040 centimeters will be stretched by a
weight of 6,346 grams, if g = 980 and k is 1.2 10 12 for copper wire.
39. The current i flowing in a series circuit with a resistance of R ohms and L
henrys t seconds after the source of electromotive force is short-circuited is
given by the relationship i = Ie~ RilL , where / is the current flowing in the
circuit before the short circuit. If i = 10 amperes, R = 0.1 ohms, t = 0.25
seconds, and L = 0.05 henrys, find /. (Take e = 2.718.)
40. If n is a positive integer, n\ has been defined as the product 1*2 n.
When n is very large, it is difficult to compute this product. However, Stirling's
formula gives (approximately) n\ = n n e~ n \S2irn. Use this formula to estimate
91, and compare the result with the true value which you should calculate
exactly. Do the same with 30!. (Hint: log (n!) = n log n - n log e +
- log 2 + - log TT + - log n).
41. If the rate of depreciation r per year is constant, the scrap value S after n years
of a machine with first cost C is given by the formula S = C(l r) n . Find the
scrap value after 10 years of a machine which originally cost $10,000, if 20
per cent per year is written off as depreciation.
5-9. CHANGE OF BASE
It is sometimes desirable to change from one logarithmic base to
another. Suppose there is available a table of logarithms to some
known base b (say 10, for example), and we wish to find the loga-
rithm of a number n to some other base a. We then let x = log& n ;
whence, by definition, n = b w . Similarly, if we let y = Iog n, then
we have n = a y .
It follows that a v = b, and our problem reduces to solving this
equation for y. Taking the logarithm of both sides to base 6, we
have
t y log& a = x log* b.
But Iog6 6 = 1. Therefore,
y = x ( : ) = log& n ( | )
J Vlogfe a/ \logb a/
/e e\
(5-5)
1 1 4 Logarithms Sec. 5-9
Example 5-15. Find log* 125, where e = 2.7183, by using a table to the base 10.
, Solution: By (5-5),
i 10K lSio 125
loge 125 = -p -
6 logic e
It is usually easier to multiply than to divide. Since division by logio e is a fairly
frequent operation in practical work, it should be noted that n xuo = 2.3026,
and the result can be obtained by multiplying by 2.3026 instead of dividing by
0.4343. Thus,
log. 125 = (2.0969) (2.3026) = 4.828.
EXERCISE 5-5
Find each of the following logarithms by using a table of common logarithms:
1. log, 10. 2. log fl 100. 3. Iog2 e. 4. log e TT.
5. log, e. 6. log* 10. 7. Iog 2 64. 8. Iog 20 1000.
9. Iog2o 100. 10. logioo 64. 11. log, 8. 12. logo.i 50.
13. Iogi 25 1000. 14. log* 20. 15. logo. 02 0.04. 16. logiooo 100.
O Right Triangles and Vectors
6-1. ROUNDING OFF NUMBERS
Numbers that arise in the applications of trigonometry are
usually not exact, but are sufficiently accurate for a given purpose.
Numbers of this kind are called approximate numbers, and the
degree of accuracy of such a number is indicated by how many
significant figures it contains. Reading from left to right, the
significant figures in a number are the digits starting with the first
non-zero digit and ending with the last non-zero digit, unless it is
definitely specified that the zeros on the right are significant. Thus,
in the numbers 2.405, 0.002405, and 240500, the digits 2, 4, 0, and 5
are significant figures. The zeros after the 5 in 240500 may or may
not be significant figures.
When it is desired to indicate whether final zeros are significant
or not, scientific notation is often used. Thus, in 2.405 10 5 , the
last significant figure is 5 ; in 2.40500 10 5 , the final two zeros are
regarded as significant.
To round off a number in which the last desired significant figure
is in the units place or in any decimal place, drop all digits that lie
to the right of the last significant figure. It is sometimes necessary
also to increase the last digit in the retained part by 1.
If the first digit in the dropped part is less than 5, the last digit
in the retained part is left unchanged. If the first digit in the
dropped part is greater than 5 or if that digit is 5 and it is followed
by digits other than 0, the last digit in the retained part is
increased by 1. Whten the dropped part consists of the digit 5
alone or the digit 5 followed only by one or more zeros, we shall
use the following procedure as an arbitrary rule in this book: If
the last digit retained is odd, this digit is increased by 1; if it is
even, it is left unchanged. This rule, although popular, is inferior
115
1 1 6 Right Triangles and Vectors Sec. 61
to common-sense rules in many cases. For example, if .245, .165,
.485, and .725 are to be rounded off to two decimal places and then
added, it would be more sensible to round off two of the numbers
in one direction and two in the other direction.
To round off a number in which the last significant figure will lie
to the left of the units place, first drop all digits to the right of the
place occupied by the last significant figure, and replace each
dropped digit to the left of the decimal point by a zero. Also,
either leave the last digit of the retained part unchanged or
increase that digit by one, in accordance with the directions just
given for dropping only a decimal part. For example, if 2533.62 is
to be rounded off to three significant figures, the result is 2530 ; and
if 487,569 is to be rounded off to three significant figures, the result
is 488,000.
There are two rules that are generally adopted by computers in
working with approximate numbers in order to guard against
retaining figures that may indicate a false degree of accuracy :
1. In adding or subtracting approximate numbers, round off the
answer in the first place at the right in which any one of the
given numbers ends.
2. In multiplying or dividing approximate numbers, round off
the answer to the fewest significant figures found in any of
the given numbers. The numbers entering a problem involv-
ing multiplication or division may be rounded off before the
computation is begun. If these numbers are rounded off,
they should have one more significant figure than the answer
is to have.
While these rules point in the right direction, it should be men-
tioned that rounding off computed quantities to as many significant
figures as there are 'in the given numbers does not necessarily
produce the degree of accuracy implied by the results. The subject
of accuracy of computation with approximate numbers is somewhat
complicated and beyond the scope of a book at this level.
6-2. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES
One of the simplest, yet important, applications of trigonometry
is in the solution of right triangles. A right triangle has, in addi-
tion to the 90 angle, five other parts. These are two acute angles
and three sides. If we know the length of any two sides, or either
acute angle and any one side, the triangle can be solved; that is,
the unknown parts can be found.
Sec. 63 Right Triangles and Vectors 1 1 7
To solve problems involving parts of tri- ~
angles, we shall find it helpful to be able to A
express the trigonometric functions of an
acute angle A of a right triangle ABC in
terms of the sides of that right triangle. To
derive suitable relationships, let us place the
acute angle in standard position, as shown in
Fig. 6-1. FIG - ^
In the right triangle ABC the side AC, which is the abscissa of
the point J5, becomes the side adjacent to the angle A ; the side CB,
which is the ordinate of B, becomes the side opposite to angle A ;
and the side AB, which is the radius vector to B, becomes the
hypotenuse. If we let the lengths of the side adjacent, the side
opposite, and the hypotenuse be represented by the symbols 6, a,
and c, respectively, we may express the six functions of the acute
angle A in terms of a, b, and c as follows :
, n + . A side opposite a
(6-1) sm A = -r r^ = - >
' hypotenuse c
, ^. A Side adjacent b
(6-2) cos A = -r r = ~ '
v ' hypotenuse c
/c ox , A side opposite a
(6-3) tan A = -7-5 ? = T >
side adjacent o
/a .. A hypotenuse c
(6-4) esc A = -~- r = - >
J side opposite a
, r A hypotenuse c
(6-5; sec A = .^ ,. = T
side adjacent 6
ff > ^ , A side adjacent b
(6-6) cot A = -r-; . = -
side opposite a
By using (6-1) to (6-6), we can express the trigonometric func-
tions of an angle of a right triangle without reference to any
coordinate system, since the ratios of the sides remain the same
regardless of the position of the triangle.
6-3. PROCEDURES FOR SOLVING RIGHT TRIANGLES
When solving a right triangle in which two parts are known, it
is advisable to arrange the work systematically and to follow A
definite procedure consisting of the following steps :
1. Draw a figure reasonably close to scale, and indicate the
known parts.
2. Write an expression containing a trigonometric function
which involves the two known parts and one unknown part.
a -658
1 1 8 Right Triangles and Vectors Sec. 6-3
3. Find the selected unknown part from this equation.
4. Find all other unknown parts of the triangle by a similar
procedure.
5. Check all results.
Whenever possible, select a trigonometric function that gives a
solution by means of a multiplication rather than a division.
In the following illustrative examples,
the acute angles are represented by the
letters A and 5, and the right angle is
denoted by C, while the small letters a,
6, and c, respectively, represent the sides
opposite them. FlG> 6 _ 2<
Example 6-1. Solve the triangle ABC, if A = 2820' and a = 658.
Solution: The triangle is drawn approximately to scale in Fig. 6-2. The unknown
parts are the angle B and the sides b and c.
Since A +B = 90, we have B = 90 - 2S20' = 6140'.
To find the side b, we may apply either (6-3) or (6-6), since both equations
involve the unknown b and the known parts A and a. We shall use cot A = -
because it enables us to proceed to the solution by means of a multiplication rather
than a division. Since A = 2S20' and a = 658, we have
Then
b = 658 cot 2820 /
= (658) (1.855) = 1220.59.
In this example, we take b equal to 1221. This result is rounded off to four digits.
**
To find the side c, we shall use (6-4), or esc A = - We have, therefore,
CSC 28020' rrJjg.
Hence,
c = 658 esc 2820'
= (658) (2.107) = 1386.
1221
To check, we may use the relation cos A = r^^ = 0.8802. Hence, A = 2820'.
loot)
Checking by means of the Pythagorean theorem yields the result b 2 = c 2 - a 2
= (c - a) (c + a) = (728) (2044) = 1488232, whereas 6 2 = (1221) 2 = 1490841.
These values of b 2 agree when they are rounded off to three significant figures,
Note. In most situations where we must solve triangles, we are
dealing with measured quantities, which are necessarily approxi-
mate. Therefore, our answers can be no more accurate than the
See. 6-3 Right Triangles and Vectors 1 19
data we begin with. If the original data are
approximate, our answers must be rounded off to
the degree of accuracy indicated by the data.
In example 6-1, for instance, the answers may
be given as b = 1220 and c = 1390, both rounded ,
off to three significant figures.
FIG. 6-3.
Example 6-2. Solve the triangle ABC, if b = 250 and c = 371.
Solution: The conditions are shown in Fig. 6-3. Since - = cos A, we have cos A
OKf| C
= j = 0.6738. Therefore, A = 4738 / and B = 90 - 4738' = 4222'.
o/ I
To find a, we have a choice of combining the unknown a with either 6 or c;
hence, we may use either -r = tan A or - = sin A . We shall illustrate, in order,
o c
the computation with each of these equations, thus providing a check on our work.
Using (6-3), we have
^= tan 47=38'.
Hence, a = 250 tan 4738' = (250) (1.096) = 274.0, or 274 when rounded off to
three figures.
Using (6-1), we have
371 = sin 47 38 ''
Hence, a = 371 sin 4738' = (371) (0.7388) = 274.09 or 274 when rounded off to
three figures.
EXERCISE 6-1
In each of the problems from 1 to 16, solve the right triangle.
1. a = 12, A - 33. 2. b = 168, A = 3816'.
3. b = 62.4, B = 7110'. 4. a = 42, c = 76.
5. o = 3.187, 6 = 6.249. 6. 6 = 63.21, B = 8336'.
7. a = 4.318, B = 6716'. 8. b = 827.6, c = 963.4.
9. a = 9.863, A = 3621'. 10. 6 = 16.32, B = 8710 / .
11. b = 78.21, A = 4317'. 12. a = 43.21, c = 63.75.
13. a = 123.6, b = 783.1. 14. a = 36.83, A = 5744'.
15. 6 = 2.312, B = 4057'. 16. a = 389.3, 6 = 62.34. ?
17. A wire stretches frorn. a point on level ground to the top of a vertical pole. It
touches the ground at a point 15 feet from the foot of the pole and makes an
angle of 63 with the horizontal. Find the height of the pole and the length
of the wire.
18. A ladder 40 feet long rests against a vertical wall. If its footjs 5 feet from the
base of the wall, what angle does it make with the ground?
120
Right Triangles and Vectors
Sec. 6-3
19. A ladder 65 feet long is placed so that it will reach a window 35 feet above the
ground on one side of a street. If the foot of the ladder is held in the same
position and the top is moved to the other side of the street, it will reach a
window 28 feet above the ground. How wide is the street from building to
building?
20. The grade of a hill is the tangent of the angle the hill makes with the horizontal.
Find the grade of a hill which is 275 feet long and which rises 120 feet.
21. To find the width of a river, a surveyor sights on a line across the river between
two points A and B on opposite banks of the river. He then runs a line AC
perpendicular to AB. He finds that AC is 250 feet and angle ACB is 4217'.
How wide is the river?
22. Find the length of a side of a regular hexagon and the radius of the inscribed
circle, if the radius of the circumscribed circle is 10 feet,
23. An airplane rises 560 feet while flying upward for 2,387 feet along an inclined
straight-line path. What is the angle of climb?
24. A pendulum 4.5 inches in length swings through an arc of 28. How high does
the bob rise above its lowest position?
25. A man 6 feet tall is walking along a straight horizontal path directly away from
a lamp post 10.5 feet high. How far is he from the post at a certain instant
when his shadow is 5 feet long?
Horizontal
"""*,,
Horizontal
B
FIG. 6-4.
6-4. ANGLES OF ELEVATION AND DEPRESSION
Let the line AB in Fig. 6-4 be a level or horizontal line, and let
an observer at the point A see an object at the point C. If the
object C is above the horizontal line AB, then the angle BAG meas-
ured up from the horizontal to the line of sight AC is called the
angle of elevation to C from A. If the object C is below the hori-
zontal line AB, then the angle BAG measured down from the'
horizontal to the line of sight AC is called the angle of depression
to C from A.
Example 6-3. From a point on the ground 300 feet from the base of a building,
the angle of elevation to its top is 22 10'. How high is the building?
a
Solution: In Fig. 6-5, we have b = 300 and A = 2210'. By (6-3),
300
= tan 2210'. Hence, a = 300 tan 2210' = (300) (0.4074) = 122.22. So the build-
ing is 122 feet high.
Sec. 6-5
\22IQ'
FIG. 6-5.
Right Triangles and Vectors
121
W-
30 B
S
FIG. 6-7.
6-5. BEARING IN NAVIGATION AND SURVEYING
In marine and air navigation and in surveying, the direction in
which an object is seen is expressed by the bearing or azimuth of
the line of sight from the observer. The bearing of a line is the
acute angle which its direction makes with a meridian or north-
south line. Such angles are sometimes called quadrant angles, or
quadrant bearings. To describe the bearing of a given direction,
we first write the letter N or S, then the acute angle, and finally
the letter E or W. The letters depend on the quadrant in which
the given direction falls. Thus, the bearings of the lines OA, OB,
and OC in Fig. 6-6 are N 60 E, S 37 W, and N 45 W, respectively.
The azimuth of a line differs from its bearing only in that the
azimuth is the angle measured from at north in a clockwise
direction. An azimuth may have any value between and 360.
Thus, in Fig. 6-6 the azimuths of the lines OA, OB, and OC are
60, 217, and 315, respectively, measured clockwise from the
north. This method of measuring directions is coming into more
frequent use than that of quadrant bearings. We note also that the
term bearing is often used instead of azimuth. Thus, we may speak
of the bearing of an object regardless of whether we mean azimuth
or bearing as here defined.
Example 6-4. A ship heads due east from a dock at a speed of 18 miles per hour.
After traveling 30 miles it turns due south and continues at the same speed. Find
its distance and bearing from the dock after 4 hours. *
Solution: In Fig. 6-7, let A be the point at which the dock is located, let B be the
point where the ship turns south, and let C be the position of the ship after 4 hours.
orj f*
Since the number of hours required to travel from A to B is ^5 = ~ > the number
lo o
5 7
of hours spent in travel from B to C is 4 - ^ = ^
Hence, a =' 18 ^ = 42.
o
Right Triangles and Vectors
42
30
and sec 6 = ~~
B = 5428',
1 22 Riaht Trianales and Vectors Sec. 6-5
From the figure,
tan0 =
Therefore,
and
b = 30 sec 5428' = (30) (1.721) = 51.6.
Hence, the distance from the dock is 52 miles and the bearing of the line AC is
14428' or S 3532' E.
6-6. PROJECTIONS
Often it is desirable to consider direction along a line segment.
Thus, if Pi and P 2 are the end points of a segment, we shall under-
stand PiP 2 to mean the directed segment from PI to P 2 , the direc-
tion being specified by the order in which the end points are named.
The non-negative length of the segment PiP 2 is denoted by |PiP 2 |.
Frequently a directed segment PiP 2 may lie on a line, such as a
coordinate axis, on which a positive direction has been specified.
Then the positive direction on the line may agree with the direction
from PI to P 2 , or the two directions may be opposite to each other.
The directed length of the segment PiP 2 is equal to |PiP 2 | when the
directions agree or when PI and P 2 coincide and is equal to -|PiP 2 |
when they disagree. Since the context will make the meaning clear,
we shall designate the directed length of the segment PiP 2 also
by PiP 2 .
M
B
FIG. 6-8.
FIG. 6-9.
We recall that the projection of a point on a given line is the foot
of the perpendicular dropped from the point to the line. If A in
Fig. 6-8 is the projection of PI on the line I, and if B is the projec-
tion of P 2 on I, then the directed segment from A to B is the
projection on I of the directed segment PjP 2 . We draw PiM parallel
to I, or perpendicular to P 2 , to show the angle between I and
Sec. 6-6
Right Triangles and Vectors
123
We shall now assume that a positive direction has been specified
on the line L Then, since P\M = AB, it follows immediately from
trigonometry that
AB = | PiP 2 | cos 0,
where is the acute angle between the positive end of I and the
positive half -line determined by the directed segment P\P* In Fig.
6-8 it is considered that I is positively directed toward the right.
The result just given can be applied, as seen in Fig. 6-9, in find-
ing the projections of PiP 2 upon the coordinate axes. The directed
lengths of the projections upon the cc-axis and the 2/-axis are,
respectively,
(6-7) AB = | PiP 2 | cos 0,
(6-8) CD = | PiP 2 | sin 6,
where 6 is the angle between OX and PiP 2 , as shown in Fig. 6-9.
When the coordinates of the end points of the segment PiP 2 are
known, the projections AB and CD are readily expressed in terms
of these coordinates. From the definitions of horizontal and vertical
distances given in Section 2-2, it follows that
(6-9) AB = x 2 xi and CD = y* - yi.
C (0,2)
0(0, -5)
P,(3,2)
A (3,0)
B(7,Q)
\67
P 2 (7,-5)
FIG. 6-10.
FIG. 6-11.
Example 6-5. What jare the projections of the segment PiPa on the axes, if
Pi = (3, 2) and P 2 = (7, - 5)?
Solution: In Fig. 6-10, AB = x 2 - %i = 7 - 3 = 4, and CD = t/ 2 - 2/i
= -5-2=-7. Since AB is + 4, we know that AB is directed to the right.
Also, since CD = - 7, we know that CD is directed downward.
1 24 Right Triangles and Vectors Sec. 6-6
Example 6-6. A ladder 12 feet long leans against the side of a house and makes
an angle of 67 with ground. Find its projections on the ground and on the side
of the house.
Solution: The conditions are represented in Fig. 6-11. Let I = 12 be the length
of the ladder. The required projections are found as follows.
The projection on the ground is given by
x = I cos 6 = 12 cos 67' = 12 (0.3907) = 4.7.
The projection on the side of the house is given by
y = I sin = 12 sin 67 = 12 (0.9205) = 11.0.
EXERCISE 6-2
1. Two points A and B are 5,000 feet apart and at the same elevation. An airplane
is 10,000 feet directly above point A. Find the angle of depression from a
horizontal line through the airplane to point B and the airplane's distance
from point B.
2. The Washington monument is approximately 555 feet high. Find the angle of
elevation to the top of the monument from a point that is 621 feet from the
base of the monument and at the same elevation as the base.
3. If a kite is 130 feet above the ground and 150 feet of string is out, find the angle
of elevation to the kite, assuming the string to He on a straight line.
4. Find the angle of elevation to the sun if a flagpole 95 feet high casts a shadow
63 feet long on horizontal ground.
5. A boat leaves its dock and heads N 52 W for 4 hours at 14 knots (1 knot = 1
nautical mile per hour = 6,080.4 feet per hour). It then turns and heads
N 38 E for 3 hours at 16 knots. Find the boat's final bearing and distance
from the dock.
6. The grade of a certain railroad bod is 0.1095. How many feet does a locomotive
rise while traveling 175 feet along the track?
7. An approach must be built up to the end of a bridge which is 40 feet above
ground. If the approach is to have a 10% grade and the original ground is
assumed to be level, how far from the end of the bridge must the approach start?
8. A surveyor wishes to find the distance between two points A and B separated
by a lake. He finds a point C on the shore of the lake such that angle ACB is
90. He measures AC and BC and finds that AC is 640 feet and BC is 285 feet.
How far apart are A and B?
9. A smokestack is 175 feet from a building. From a window of the building the
angle of elevation to the top of the stack is 2810 / . The angle of depression to
its base from the same window is 2430'. Assuming that the ground is level,
find a) the height of the window above the ground and b) the height of the
smokestack.
10. Two ships leave the same port at the same time. One travels N 42 E at 25
knots. The other travels S 48 E at 33 knots. How far apart are the two ships
after 4 hours?
Sec. 6-7
Right Triangles and Vectors
125
6-7. SCALAR AND VECTOR QUANTITIES
We shall at this point find it necessary to distinguish carefully
between two kinds of quantities, namely scalar quantities and vector
quantities.
A scalar quantity is a quantity whose measure can be fully
described by a number. It is a quantity which can be measured on
a real number scale. For example, temperature is a scalar quan-
tity, measured on the scale of a thermometer. Also we shall define
the scalar components of the segment PiP 2 to be the projections on
the coordinate axes, or the directed lengths x 2 Xi and y 2 3/1
given by (6-9) in Section 6-6. The student should note, however,
that the scalar components of P 2 Pi are not equal to those of P\P 2 .
The components of PJP\ are x\ x 2 and yi y 2 and are the nega-
tives of the respective components of PiP 2 .
A vector quantity, or simply a vector, is a quantity possessing
both magnitude and direction. A vector may be represented by any
one of a set of equal and parallel line segments. Algebraically,
a vector is fully described by the scalar components of any segment
representing it ; all such segments have the same components. We
shall, in fact, call these the scalar components of the vector and shall
./
o
FIG. 6-12.
FIG. 6-13.
enclose them in brackets. In Fig. 6-12, three representations of the
vector [4, 3] are shown. The arrowhead indicates the order in
which the end points of the line segment are named.
We may denote a vector by a single letter in boldface type, say v,
or may represent it geometrically by any one of the segments, such as
AB, OP, or CD. Here A, O, and C represent the initial points of the
three segments, while the terminal points at the arrowheads are
B, P, and D. Actually, any other segment with the same magnitude
and direction could have been selected to represent the vector v.
Instead of using a single letter in boldface type to denote a vector
represented by a segment PiP 2 , we may also use the notation PiP2.
1 26 Right Triangles and Vectors Sec. 6-7
Properties of Vectors. We have seen that a vector is unambigu-
ously represented by any one of a set of equal and parallel line
segments. Hence, any point may be taken as the initial point
of a segment which represents a vector. If the origin is so chosen,
the coordinates of the end^point P(x,y) are actually the scalar
components of the vector OP. We can therefore give the following
simple definition of the magnitude of a vector and its expression
in terms of scalar components :
Definition. The magnitude, or length, of a vector v is the length of
any one of the segments representing v.
Let v = [yi, v 2 ] be the vector represented by OP in Fig. 6-13,
where v\ and v 2 are scalar components. Then we have |v| = \OP\.
Since OP is the hypotenuse of a right triangle,
(6-10) | v | = VV + v 2 2 .
In case the scalar components are given by the coordinates (x, y)
of the end point P of the segment, the magnitude of the vector is
From the definition of a vector, it follows that two vectors are
equal if and only if their respective scalar components are equal.
For example, u = [u lt u 2 ] equals v = |>i, v 2 ] if and only if HI = Vi
and u 2 = v 2 .
We also define a special vector = [0, 0] to be the zero vector.
It corresponds to the exceptional case in which P 2 coincides with
PI and may be considered as represented geometrically by a seg-
ment of length zero, that is, by a point. The zero vector may be
regarded as having any direction whatsoever.
If a vector v. = [v lf v 2 ] is given, the vector v = [ Vi, v 2 ] is
defined to be the negative of v. Thus, ifj^ift = v denotes a vector
represented by the segment PiP 2 , then P 2 Pi = v denotes a vector
having the same length as PiP 2 but oppositely directed, namely,
from P 2 to PI. We note that
- (- Y) = v.
A unit vector is defined as a vector whose magnitude is unity.
If u = [fa, u^ is any non-zero vector, then -, = \T^I > r^]\
is a unit vector.
Multiplication of a vector by a scalar is performed by multiplying
the magnitude of the vector by the absolute value of the scalar,
maintaining the direction of the vector if the scalar is non-negative
and reversing it otherwise. Thus, by k[ui, u^] we mean the vector
Sec. 6-7 Ri ghf Triangles and Vectors 1 27
[kui, ku2], Using the equation [MI, u 2 ] = k [~ ? yl i we can express
any vector v = Oi,^ 2 ] as proportional to a unit vector v/fc if we
choose A: = V^i 2 + ^2 2 - Changing v to a unit vector v/fc is called
normalizing v.
Sums and Differences of Vectors. Of the many questions which
arise in the study of vectors, the one of greatest importance for us
at present concerns the addition of vectors. The sum, or resultant,
of two vectors is defined to be the vector which has for its scalar
components the sums of the scalar components of the two vectors.
Thus, the sum of the vectors u and v is given by the relationship
(6-11) u + v = [MI + vi, u 2 + v 2 ].
It is important to note that some of the laws of the algebra of
numbers also hold for vectors. Thus, the commutative law is
u + v = v + u.
Also, the associative law of addition is
(u + v) + w = u + (v + w).
And, for every vector v,
v + ( - v) = 0.
That these laws are satisfied for vectors can be shown geometrically
or can be seen from (6-11) by virtue of the known laws of addition
of real numbers.
In terms of components, the rule for multiplication by a scalar is
given by
In consequence of this definition, the following algebraic laws are
satisfied :
fc(u + v) = fcu + &v;
(k + m)u = fcu + mu;
fc(mu) = (fcw)u.
To find the sum of two vectors geometrically, we proceed as
indicated in Fig. 6-14. This graphical representation of the sum
of two vectors by means of the triangle construction was probably
suggested by the behavior of physical quantities represented by
vectors, such as forces, displacements, velocities, and accelerations.
Their addition is effected by the triangle law or the parallelogram
law.
128
Right Triangles and Vectors
Sec. 6-7
o
1C
FIG. 6-14.
FIG. 6-15.
2.
3.
The following steps indicate the actual procedure.
1. Select a segment representing one of the vectors, say u, with
its initial point at the origin of the coordinate system.
Place the initial point of a segment representing the second
vector, say v, at the terminal point of the first segment.
Draw the segment from the origin to the terminal point of the
segment for v. This segment represents the resultant vector,
u 4- v.
The difference u v of two vectors is defined in a manner
analogous to that used in defining the difference of numbers. Thus,
u v = u + ( v).
If u = [ui, u%~\ and v [y\, v%] , then
U V = [U\ Vi, U2 V2\.
To find the difference u v of two vectors geometrically, we
proceed as indicated in Fig. 6-15. The steps are as follows :
1. Place segments representing u and v with their initial points
at the origin.
2. Place the initial point of a segment representing v at the
terminal point of the segment representing u. Note that the
segment representing v is parallel and equal in length to
that for v, but has the opposite direction.
3. Draw the segment from the origin to the terminal point of
the segment for v. This segment represents the vector
u v.
Example 6-7. Find the sum and difference of the vectors u = [5, - 3] and
Solution: The required vectors are
u+ v = [5 - 2, - 3 + 1] = [3, - 2],
and
u-? = |5-(-2), -3-l] = [7, -4].
Sec. 6-7
Right Triangles and Vectors
129
Example 6-8. Express u = [4, 3] as proportional to a unit vector.
Solution: The expression representing the vector is
[A Q~1
- > - . To check,
note that the magnitude of -r > -r is
to*)
To add vectors analytically, we first find the scalar components
of the vectors to be added.
From Section 6-7, we know
that if segments represent-
ing u and v make angles a
and /?, respectively, with
the #-axis, the scalar com-
ponents are given by the
projections upon the coordi-
nate axes. Thus, as shown
in Fig. 6-16,
and
v x = | v | cos ]8 ; vy = | v | sin ]8.
Then the components of the resultant will be given by the
algebraic sums
X = U x + V x , Y = Uv + Vy.
Thus, the magnitude of the resultant R is given by
|R| =
u x = | u | cos a t
FIG. 6-16.
= I u I sin a,
Also, the direction angle 6 satisfies the relationship
Y
tan 6 = j=
To determine the quadrant of correctly, it is important to keep
in mind and use the correct signs of X and Y. For example, tan
= 1 might lead one to an incorrect value 45 for 6 instead of
- 1
the correct third-quadrant angle 225.
The student should note that this (analytic) procedure and the
geometric procedure (parallelogram law) give the same results. This
can be shown by appropriately combining the scalar components of
the vectors u, v, u + v, and u - v in Figs. 6-14 and 6-15, where we
illustrated the geometric procedure.
130
Right Triangles and Vectors
Sec. 6-7
W-
\25 C
D
FIG. 6-17.
o
o
^*
\
s
FIG. 6-18.
-*-E
Example 6-9. A block weighing 500 pounds rests on a smooth plane making an
angle of 25 with the horizontal, as indicated in Fig. 6-17. What force, parallel to
the plane, is necessary to hold the block in position?
Solution: Let the block be at A. The force due to the weight mayjbe represented
by a vector acting vertically downward through A, such as vector AC. The com-
ponent of AC which is parallel to the inclined plane and which must be overcome is
represented by the projection A B of AC on the inclined plane. Since angle
BAG = 90 - 25 = 65,
AB = 500 cos 65 = (500) (0.4226) = 211.3.
Hence, a force of 211 pounds must be applied parallel to the inclined plane to keep
the block from sliding. (It is assumed that three significant figures are appropriate.)
Example 6-10. Find the magnitude and direction of the resultant of a force of
110 pounds acting in the direction S 4627' E and a force of 100 pounds acting in
the direction N 2914' W.
Solution: The given forces are represented by vectors in Fig. 6-18. Let F denote
the magnitude of the resultant force and 6 the angle it makes with OE.
Since 90 - 4627' = 4333' and 90 - 2914' = 6046', the components F x and
F y of the resultant are found as follows:
F x = 110 cos 4333' - 100 cos 6046'
= (110) (0.7248) - (100) (0.4884)
= 79.73 -48.84=30.89;
F v = - 110 sin 4333' + 100 sin 6046'
= - (110) (0.6890) + (100) (0.8726)
= - 75.79 + 87.26 = 11.47.
Now
tan*=^ =
11.47
= 0.3713, and esc 6 =
F
30.89 ~~ v '"' *"' """ ^ v ~~ 11.47
Therefore B = 2022', and F = 11.47 esc 2022' = (11.47) (2.873) = 32.95. Hence,
the magnitude of F is 33 pounds, and its direction is N 6938' E.
Sec. 6-8 Right Triangles and Vectors 131
EXERCISE 6-3
1. Draw a diagram showing three different segments representing each of the
given vectors. Find the magnitude of each vector.
a. [3, 4]. b. [12, 5]. c. [-2, 4]. d. [- 3, - 5].
e. [1, - 1]. f. [5, 3]. g. [0, 4]. h. [0, - 1].
2. Express each of the vectors in Problem 1 in terms of a unit vector.
3. In each of the following cases, add the given vectors. Find the magnitude and
direction of the resultant.
a. [1, 1] and [2, 3]. b. [4, - 2] and [1, - 10].
c. [2, 0] and [- 6, 3]. d. [1, 2], [4, 3], and [0, 7].
4. In Problem 3, parts (a), (6), and (c), subtract the first vector from the second,
and find the magnitude and direction of the resultant.
5. In Problem 3(d), subtract twice the third vector from the sum of four times the
first vector and twice the second vector, and find the magnitude and direction
of the resultant.
6. A force of 40 pounds acts at an angle of 63 with the horizontal. What are the
vertical and horizontal components of the force?
7. If a ship sails N 48 W at 30 knots, what are its westward and northward
components?
8. One force of 28 pounds acts vertically upward on a particle. Another force of
43 pounds acts horizontally on the particle. What is the magnitude of the
resultant force, and what is its direction?
9. A barrel weighing 160 pounds rests on a smooth plane which makes an angle
of 22 with the horizontal. Find the force parallel to the plane necessary to
keep the barrel from rolling down the plane.
10. Three forces act on a particle. One of 50 pounds makes an angle of 25 with the
horizontal; a second of 60 pounds makes an angle of 50 with the horizontal;
and the third of 75 pounds makes an angle of 230 with the horizontal. Find
the magnitude and direction of the resultant force.
11. Four forces act on a body. The forces are 30, 45, 50, and 65 pounds, and they
make angles with the horizontal of 25, 160, 240, and 330, respectively.
Assuming that all the forces lie in the same vertical plane, find the magnitude
and direction of the force necessary to hold the body in equilibrium. The
required force is equal in magnitude and opposite to the resultant of the given
forces.
12. A shell is fired at an angle of elevation of 37. Its initial velocity is 2,500 feet
per second. Find the horizontal and vertical components of its initial velocity.
6-8. LOGARITHMS OP TRIGONOMETRIC FUNCTIONS
So far in this chapter, we have considered the use of a table Qf
natural trigonometric functions and have solved various problems
involving right triangles. In many problems, however, the compu-
tation is greatly facilitated by the use of logarithms to perform the
numerical operations. For this purpose the values of the logarithms
of the trigonometric functions are required. Table III might be
used to obtain the logarithms of the functions found in Table II,
132
Right Triangles and Vectors
Sec. 6-8
but the work is considerably lessened by the use of Table IV at the
end of this book, which gives the logarithms of the trigonometric
functions at once.
Table IV is a four-place table giving the logarithms of functions
at intervals of 10 minutes from to 90, For the sine and cosine
of any angle between and 90, the tangent of any angle between
and 45, and the cotangent of any angle between 45 and 90,
the value of the function is less than 1; hence, the logarithms of
these functions are negative, and we must write 10 after the
tabulated entry. For the sake of uniformity, 10 has been added
to each of the other entries in the table. In using the table,
therefore, 10 must be subtracted from every entry.
The method of using Table IV is similar to that described for
Table II and will be illustrated by the following examples.
Example 6-11. Find log sin 2310'.
Solution: Since this angle is given in Table IV, we find that log sin 23 10'
= 9.5948-10.
Example 6-12. Find log cot 5127'.
Solution: From Table IV we obtain the values for the following tabulation:
log cot 5120' - 9.9032 - 10
10'
log cot 5127 ; = 9.9032 - 10 - x
log cot 5130' = 9.9006 - 10
26
The tabular difference is 26. Since 5127' is -^ of the way from 5120' to 5130',
x = 0.7(26) = 18.2, and we have
log cot 5127' = (9.9032-10) - .0018
= 9.9014-10.
Example 6-13. Find the acute angle 6 if log tan 6 = 9.7827-10.
Solution: The positive part 9.7827 lies between the entries 9.7816 and 9.7845
in Table IV. The procedure for finding 6 may be indicated as follows:
log tan 3110' = 9.7816 - 10
10'
TT x II .
Hence, = > and
log tan 31(10 + x) f = 9.7827 - 10
log tan 3120' = 9.7845 - 10
8 = 3110' + 55(10')
11
29
Sec. 6-9 Right triangles and Vectors 1 33
EXERCISE 6-4
In each of the problems from 1 to 15, find the value of the given logarithm.
1. log sin 4820'. 2, log sin 2120'. 3. log cos 8620'.
4. log tan 8830'. 5. log cot 1020'. 6. log sec 4350'.
7. log sin 1326'. 8. log sec 4857'. 9. log tan 4114'.
10. log esc 7832'. 11. log cot 6843'. 12. log cos 1818'.
13. log esc 8316'. 14. log cos 18'. 15. log tan 5134 / .
In each of the problems from 16 to 35, find the angle (or angles) 6 between
and 360.
16. log sin 6 = 8.8059-10. 17. log tan 6 = 8.3661MO,
18. log cos 6 = 9.9959-10. 19. log sin 6 = 9.1697-10.
20. log sec = 0.4625. 21. log cot 6 = 0.4882.
22. log tan 9 = 9.8483-10. 23. log tan 9 = 0.1430.
24. log esc 6 = 0.4081. 25. log sec 6 = 0.3586.
26. log sin 9 = 9.9567-10. 27. log tan 9 = 9.7648-10.
28. log cos 9 = 9.9755-10. 29. log cot 9 = 9.8666-10.
30. log sec 9 = 0.1967. 31. log esc 6 = 0.3370.
32. log cos 9 = 9.1860-10. 33. log cot 9 = 1.5976.
34. log sin 9 = 9.9974-10. 35. log sec 9 = 0.3870.
6-9. LOGARITHMIC SOLUTION OF RIGHT TRIANGLES
The solution of a right triangle by means of logarithms is exactly
the same as by natural functions, except that
for the actual numerical computation a table
of logarithms of the natural functions is used
in conjunction with a table of logarithms of
numbers. The following example will illus-
trate the procedure.
Example 6-14. Solve the right triangle ABC, in which A = 2140' and b = 8.43.
Solution: The values of the known parts are indicated in Fig. 6-19. We see that
B = 90 - 2140' = 6820'. To find the side a, we have
tan 2140' = ^ >
8.4o
Hence,
a = 8.43 tan 2140',
or
log a = log 8.43 -f log tan 2140'.
Arrange the work as follows:
log 8.43 = 0.9258 (Table III)
. log tan 2140' = 9.5991-10 (Table IV)
log a = 10.5249-10
Therefore, a = 3.35. (Table III)
To find side c, we have
= cos 2140'.
c
1 34 Right Triangles and Vectors Sec. 6-9
Hence, 8 . 43
C cos 2140' '
and
log c = log 8.43 - Jog cos 2140'.
Arrange the work as follows:
log 8.43 = 10.9258-10 (Table III)
log cos 2140' = 9.9682-10 (Table IV)
log c = 0.9576
Therefore, c = 9.07. (Table III)
EXERCISE 6-5
In each of the problems from 1 to 8, solve the given right triangle.
I. 6 = 100, A = 31. 2. c = 3.45, a = 1.76. 3. A = 2520', a = 63.4.
4. A = 8817', c = 108.1. 5. c = 6.275, B = 1845'. 6. a = 645.3, b = 396.3.
7. J3 = 279', a = 36.13. 8. 6 = 98.34, B = 1848'.
9. If a railroad track rises 30 feet in a horizontal distance of one mile, find the
angle of inclination of the track.
10. A force of 341 pounds and another force of 427 pounds act at right angles to
each other. Find the magnitude of the resultant force and the angle it makes
with each of the forces.
II. A force of 628 pounds acts at 180, and a force of 237 pounds acts at 270. Find
the direction and magnitude of the resultant.
12. An airplane is flying due east at a speed of 485 miles per hour, and the wind
is blowing due south at 33.6 miles per hour. Find the direction and speed
of the phane.
13. The westward and northward components of the velocity of an airplane are
363 and 487 miles, respectively. Find the direction and speed of the airplane.
14. The eastward and southward components of the velocity of a ship are 10.4 and
16.8 knots, respectively. Find the speed of the ship and the direction in which
it is moving.
15. A force of 2673 pounds is acting at an angle of 4713' with the horizontal.
Find its horizontal and vertical components.
16. A force of 162.4 pounds is just sufficient to keep a block at rest on a smooth
inclined plane. If the block weighs* 783.1 pounds, find the angle at which the
plane is inclined to the horizontal.
17. Two tangents are drawn from a point P to a circle whose radius is 14,32 inches.
If the angle between the tangents is 3228', how long is each tangent segment?
18. Two buildings of the same height are 11,640 feet apart. When an airplane is
8,000 feet above one of them, what is the angle of depression to the other one?
19. A cable which can withstand a pull of 10,000 pounds is used to pull loaded
trucks up a ramp. If the angle of inclination of the ramp is 3616', find the
weight of the heaviest truck which can be safely pulled up the ramp with
the cable.
20. The angle of elevation from one point on level ground to the top of a flagpole
is 4528'. From a point on the ground 25 feet farther away the angle is 3956'.
How high is the pole?
Trigonometric Functions of
Sums and Differences
7-1. DERIVATION OF THE ADDITION FORMULAS
Heretofore, we were concerned with relationships between trig-
onometric functions of a single angle. We shall now establish
certain fundamental identities involving two angles, in terms of
the functions of the single angles. The following identities express
functions of the sum and difference of two angles in terms* of the
functions of the separate angles.
(7-1)
(7-2)
(7-3)
(7-4)
(7-5)
(7-6)
sin (a + j8) = sin a cos |8 + cos a sin /3,
cos (a + /3) = cos a cos /3 sin a sin /3,
sin (a /3) = sin a cos /3 cos a sin |8,
cos (a /3) = cos a cos ft + sin a sin j8,
, . ^
tan (a + p) =
tan
tan ]
, ^
tan (a p) =
v K/
-
1 tan a tan p
tan oj tan j8
1 + tan a tan /3
We shall now prove the formulas for the sine and cosine by using
the derivation developed by E. J. McShane. 1
o
FIG. 7-1.
FIG. 7-2.
1 E. J. McShane. "The Addition Formulas for tho Sine and Cosine,"
American Mathematical- Monthly, Vol. 48 (1941), pp. 688-89.
135
136
Trigonometric Functions of Sums & Differences
Sec. 7-1
Let and a be any two angles with the same initial side OW, as
shown in Fig. 7-1. On their terminal sides we choose points P and
Q, respectively, each at unit dis-
tance from 0.
Let d represent the distance
from P to Q. We shall now make
two computations for d 2 , using
first OW, and then OP, as the
0[cos(-0). sin (a -0)]
FIG. 7-3.
When OW is used as the x-axis
of a coordinate system, as shown
in Fig. 7-2, we find that the coordi-
nates of P and Q are (cos 0, sin /?)
and (cos a, sin a), respectively.
Hence, by the distance formula,
d 2 = (cos a cos 0) 2 + (sin a sin 0) 2
= cos 2 a 2 cos a cos + cos 2 + sin 2 a 2 sin a sin + sin 2 0.
Since cos 2 a + sin 2 a = cos 2 + sin 2 = 1 ,
d 2 = 2 2 (cos a cos + sin a sin 0).
Let us now use OP as the #-axis, as shown in Fig. 7-3. Then the
coordinates of P are (1,0) and those of Q are [cos (a /3),
sin (a 0) ] . Hence,
d 2 = [cos (a - 0) - I] 2 + sin 2 (a - 0)
= cos 2 (a j8) '2 cos (a j8) + 1 + sin 2 (a 0)
= 2 - 2 cos (a - 0).
Equating the two expressions for d 2 yields
2 2(cos a cos + sin a sin j8) =2 2 cos (a |8).
Therefore,
(7-4) cos (a 0) = cos a cos + sin a sin ]8.
This establishes (7-4).
Setting a = 90 in (7-4), we find that
(7-7) cos (90 - 0) = sin 0.
If in (7-7) we let /3 = 90 - y, we have
(7-8) sin (90 - 7) = cos 7.
From (7-7), (7-8), and (7-4), we obtain
sin (a + 0) = cos [90 - (a + 0)]
= cos [(90 - a) - 0]
= cos (90 - a) cos + sin (90 - a) sin 0,
Sec. 7-1 Trigonometric Functions of Sums & Differences 137
or
(7-1) sin (a + /3) = sin a cos /3 + cos a sin /3.
This establishes (7-1).
Since cos (-)8) = cos /3 and sin (-/?) = -sin , we have, as a con-
sequence of (7-4),
cos (a + /3) = cos [a ( |8)]
= cos a cos ( j8) + sin a sin ( j8)
or
(7-2) cos (a + ]8) = cos a cos j8 sin a sin ]8.
Similarly, from (7-1) it follows that
(7-3) sin (a: ]8) = sin a. cos ( j8) + cos a sin ( ]8)
= sin a cos |8 cos a sin /3.
To prove (7-5), we use the relationship tan 6 = sm fl and (7-1)
COS C7
and (7-2). We then obtain
, , m sin (a + fi) sin a cos B + cos a sin ]8
tan (a + p) = 7 r^\ ~ 5 '- r ~# '
cos (a + p) cos a cos p sin a sm p
Dividing each term of the numerator and denominator by
cos a cos /3, we have
sin a cos j8 _, cos a sin )8
cos a cos j8 cos a cosjg _ tan a + tan ff ^
cos a cos j8 __ sin a sin ff 1 tan a tan /3
TT cos a cos B cos a cos 8
Hence, ' , , Q
t* ^ / . /o\ tan a + tan p
(7-5) tan (a + |8) = - ^5
v y v ^ y 1 - tan a tan p
We may obtain (7-6) in a similar manner from (7-3) and (7-4),
or from (7-5).
Example 7-1. Find the exact value of sin 75.
Solution: Substituting 45 for a and 30 for in (7-1), we obtain
sin 75 = sin (45 + 30)
= sin 45 cos 30 + cos 45 sin 30
_\/2 V3 V2 1
"2*2^2*2
Therefore, .-=
sin 75 = V (V 3 + D-
4
138 Trigonometric Functions of Sums & Differences Sec. 7-1
Example 7-2, Find the exact value of tan (a. + /3) if a is a second-quadrant
3 5
angle such that sin a = = , and is a third-quadrant angle tan ft =
O \JL
Solution: Let a be a second-quadrant angle for which y = 3 and r = 5.
3
Hence, x = 4. It follows that tan a = - j ' Using (7-5), we have
x / , m ~ 4 + 12 16
tan (a + 6) = 7-^-7 = - ^
EXERCISE 7-1
In each of the problems from 1 to 8, find the exact value of the given function.
1. cos 75. 2. tan 105. 3. sin 135. 4. sin 15.
5. tan 195. 6. cos 195. 7. tan 15. 8. cos 105.
9. If a is a third-quadrant angle and /3 is a second-quadrant angle, and
3 5
sin a = - -r and cos ft = rz > find sin (a -f j(S), cos (a -f 0), and tan (a -f |8).
O lo
10. If tan a = r and a - B = 45, find tan 0.
,4
11. If tan a = 3 and a + = 180, find tan 0.
3 1
12. If tan a = j and tan j8 = - > find sin (a + j8) and cos (a + )8), where a and
4 o
jft are first-quadrant angles.
13. If cos a. = - and cos = > find sin (a j9) and cos (a j8), where a and
J o
are acute angles.
4 5
14. If sin a = -= > tan # = ^r > and a and /3 are both obtuse, find sin (a + /3)
O LZ
and cos (a -f 0).
3 24
15. If sin a = ^ > cos j8 = ^rr > a is obtuse, and j3 is acute, find sin (a 0) and
^O
cos (a j9).
2 1
16. If cos a = ~ > sin |3 = - > and a and ]S are obtuse, find sin (a -f j3) and
cos (a + |8).
3 5
17. If sin a = - > cos j8 = 7^ > a is obtuse, and is in the third quadrant, find
O Id
tan (a + ft) and tan (a - 0).
Prove each of the following identities:
A/2
18. sin (a 45) = -jp (sin a cos a). 19. cos (a TT) = - cos a.
20. Sin(a 7?=cot|8-cota. 21. tan (a +-) =i-*^.
sin a sin p \ 4 / 1 ~ tan a
22. cos 2 a = cos 2 a - sin 2 a = 2 cos 2 a - 1 = 1 - 2 sin 2 a.
Sec. 7-2 Trigonometric Functions of Sums & Differences 1 39
no o n <n si n 2 a cos 2 a
23. sin 2 a =2 sin a cos a. 24. : ----- = sec a.
sin a cos ct
sin ( a + /3) + sin (a - 0) sin (a + 0) -
25. - - -_ _ - . - ~_ = tan a, 26. ---- ~ = tan a + tan p.
cos (a + p) + cos (a - p) cos a; cos ft
97 s fo ( a + -ft) * an 0= + tan ft ^ ^ cos (a 0) _ 1 + tan a tan ft $
sin (a - ft) tan a - tan jS cos^a 4- P) ~~ I tan a tan
9Q cos ( a "" 0) - 1 + tan a tan (3 cot a cot - 1
29 ' sin (a f " l^TTtSJ ' " '
rin (a + g) sin (a /?) 2
_
cos 2 a cos 2 p
32. sin (A + B + C) = sin A cos cos (7 -f cos A sin 5 cos C
+ cos A cos B sin (7 sin A sin 5 sin C.
33. sin 3 a = sin 5 a cos 2 a cos 5 a sin 2 a.
34. cos (a - 0) cos (a 4- 0) = cos 2 a - sin 2 = cos 2 ft - sin 2 a.
35. sin 2 + ~~ cos2 a + ~ sin 2 a '
7-2. THE DOUBLE-ANGLE FORMULAS
If we let ft = a in (7-1), (7-2), and (7-5), we obtain functions
of twice a given angle in terms of the functions of the angle itself.
Thus, we have the following identities :
(7-9) sin 2 a = 2 sin a cos a,
(7-10) cos 2 a = cos 2 a sin 2 a,
and
/- * * \ n 2 tan Q!
(7-11) tan 2 a =
tan 2 a
We may obtain two other useful forms for cos 2 a from (7-10)
by using in turn cos 2 a 1 sin 2 a and sin 2 a = 1 cos 2 a. These
forms are given by the identities
(7-12) cos 2 a = 1 - 2 sin 2 a,
and
(7-13) cos 2 a = 2 cos 2 a - 1.
The following illustrations give an indication of the possible
applications of (7-9), (7-10), and (7-11). The student should
study them carefully.
sin 4 a = sin 2 (2 a) = 2 sin 2 a cos 2 a,
. /a\ . a a;
sm a = sin 2 ( ^ ) = 2 sm c <> s o '
oj rt /a;\ 9 a . 9 a
cos - = cos 2 ^ j = cos 2 - sin 2 g
1 40 Trigonometric Functions of Sums & Differences Sec. 7-2
Example 7-3. Find the exact value of sin 120 by means of a double-angle formula.
Solution: We use (7-9) to obtain
sin 120 = sin 2(60) = 2 sin 60 cos 60
\ /1\ _
V (2) -
Example 7-4. Derive a formula for cos 3 a in terms of cos a.
Solution: Applying the identity for cos (a -f /3), and the double-angle formulas,
we have
cos 3 a = cos (a + 2 a) = cos a cos 2 a sin a sin 2 a
= cos a (2 cos 2 a 1) sin a (2 sin a cos a)
= 2 cos 3 a cos a 2 sin 2 a cos a
= 2 cos a (cos 2 a; sin 2 a) cos a
= 2 cos a (2 cos 2 a 1) cos a
= 4 cos 3 a 3 cos a.
_. - -r* ,1 i , , sin 3 cos 30 _
Example 7-5. Prove the identity rz -r- 2.
r sm cos
Solution: First combine the fractions on the left side and then reduce the result
to the right side. Thus,
sin 30 __ cos_3_0 __ sin 3 cos - cos 3 sin _ sin (3 - 0)
sin cos ~" sin cos ~~ sin cos
- sin 20 _ 2_smJ!0 __ 2
~~ sin cos ~~ sin 2 "~~
7-3. THE HALF-ANGLE FORMULAS
Functions of an angle in terms of the functions of twice that
angle can be obtained directly from (7-12) and (7-13) . If cos 2 a =
1 2 sin 2 a is solved for sin a, we obtain
A /I cos 2 a
sin a = d= 4/
r ^
Also, by solving cos 2 a = 2 cos 2 a 1 for cos a, we have
, ., +cos2.
cos a = '
2
Since these formulas may be equally well regarded as expressing
functions of half an angle in terms of the functions of the given
angle itself, the same relationship is retained if the identities are
written
,~ * A \ OL A /I cos a
(7-14) sm ^ = d
and
/ 1K x v* . A / * T- cos a
(7-15) cos - = db
Sec. 7-3
Trigonometric Functions of Sums & Differences
141
These are the so-called half -angle formulas for the sine and cosine.
From (7-14) and (7-15) we obtain, by division,
, ^x
(7-16)
^ '
tan
a __ sin a/2 __ */l
2 ~~ cos a/2 ~ V I
cos a
+ cos a
The algebraic signs in (7-14), (7-15), and (7-16) are deter-
mined by the quadrant of a/2.
If we rationalize, in turn, the numerator and the denominator of
the right hand side of (7-16) , we obtain
a _ A / 1 - cos 2 a __ A / sin 2 a
tan 2 - y (1 + coga)2 - y (1 + coga)2
or
(7 17^
sin a
\< **) **" 2
Similarly,
tanf =
or
^7-1^ fon -
1 + cos a
- /(I cos a) 2 ./(I cos a) 2
- 1 - T 1 ~cos 2 a " T sin 2 a
1 cos a
r-24
The student will note that, in deriv-
ing (7-17) and (7-18), we have
dropped the sign in each formula.
The validity of this step should be
verified by consideration of the signs
of tan a/2, sin a, and 1 cos a. Thus,
tan a/2 and sin a necessarily have the
same sign, while 1 . cos a is non-
negative.
Example 7-6. Find the exact value of
tan 22.5.
Solution: Since the exact values of the functions of 45 are known, we may use
(7-16), (7-17) or (7-18). Selecting (7-18), we obtain
x rtrteo x 45 1~ cos 45
tan 22.50 = tan _ = _
V2
1 42 Trigonometric Functions of Sums & Differences Sec. 7-3
24
Example 7-7. Given tan 2 a. = - -=- > where 2 a is a second-quadrant angle.
Find sin a and cos a.
24
Solution: Since 2 a is an angle whose tangent is -=- * we may find a point on
the terminal side of the angle with x = 7 and y = 24, as shown in Fig. 7-4. Thus,
r = V49 + 576 = 25. Therefore,
,4/1+7 '25 4
B ina = >|/ 2 =5-
and
j /I - 7/25 3
cos a. l\/ ~ = -
' O
Example 7-8. Given that tan - = w, find sin a and cos a in terms of w.
Solution: Squaring both sides of (7-16), we have
2 ex __ 1 cos a
2 ~~ 1 -f cos a
Substituting u for tan - > we get u 2 = ^ If we solve this equation for
/ i , i i ^ J- i COS QJ
cos a, we find that
1 - u 2
cos a = - ; -
1 -f- u 2
If we substitute this value of cos a in the relationship sin 2 a + cos 2 a. = 1,
we obtain
-
-
Note that we can drop the i sign, just as we did in (7-17) and (7-18), since
tan a/2 and sin a always have the same sign.
EXERCISE 7-2
In each of the problems from 1 to 8, find the exact functional value by using an
appropriate double-angle or half-angle formula.
I. sin 22.5. 2. cos 15. 3. sin 120. 4. cos 90.
5. sin 67.5. 6. cos 67.5. 7. tan 67.5. 8. tan 60.
12
9. It is known that cos Q = ? and 6 is positive in the second quadrant. Find :
lo
a. sin 26. b. cos 0/2. c. tan 26. d. cot 26.
e. sin 6/2. f. cos 26. g. sin 30. h. tan 46.
40
10. It is known that sin 6 = - and 6 is positive in the third quadrant. Find :
a. sin 26. b. cos 6/2. c. tan 26. d. cot 26.
e. sin 6/2. f. cos 26. g. sin 36. h. tan 40.
II. It is known that tan 6 = - 3 and 6 is positive in the fourth quadrant. Find:
a. sin 26. b. cos 0/2. c. tan 20. d. cot 20.
e. sin 0/2. f. cos 20. g. sin 30. h. tan 40.
Sec. 7-4 Trigonometric Functions of Sums & Differences 1 43
In each of the problems from 12 to 28, write the given expression in terms of a
single function of a multiple of 6. Make use of appropriate formulas to reduce the
answer to as few terms as possible.
fi ft
12. 2 sin ^ cos jr
a i
2 tan 30
13.
16.
19.
22.
9*
C082 |_ sin2 |.
14. 2 cos 2 30 - 1.
17 -in "0 A/ 1 ~ C S 4 ^
5 " 1 - tan 2 30
"1C I Qirt fn<3 1
sm 4
cos^ sin'* G
17. sin 20 y 2
OA sin 20 cos 20
\ i /
-1 2
2 cote
ZU. . /> ' ' / *
sm cos
1 - tan 2 0/2
" * cot - tan
/9
91 rrf -4- tan
I + cot 2
sin 0/2
1 + tan 2 0/2
2 tan 0/2
Z t
1 - cos 30
1 - cos 0/2
nr\o2 (A2\ o\v\2 ffi2\
1 + tan 2 0/2
Prove each of the following identities:
29. sin 40-4 cos 0(sin 0-2 sin 3 0), 30. cos 40 = 8 cos 4 0-8 cos 2 9+1.
31. sin 20 = (1 + cos 20) tan 0. 32. 1 + sin 20 = (sin + cos 0) 2 .
^ 1- tan 2 0/2 = 2(1- cos 0). 4 sin cos 2 = sin + sin 30.
cos sin 2
7-4. PRODUCTS OF TWO FUNCTIONS EXPRESSED AS SUMS, AND SUMS
EXPRESSED AS PRODUCTS
By adding and subtracting corresponding members of (7-1),
(7-2), (7-3), and (7-4), we obtain
(7-19) sin (a + ]8) + sin (a - j8) = 2 sin a cos j8,
(7-20) sin (a + /8) sin (a - |8) = 2 cos a sin |8,
(7-21) cos (a + ]8) + cos (a. - j8) = 2 cos a cos ]8,
(7-22) cos (a + j8) - cos (a - ]8) = - 2 sin a sin /3.
If we reverse these identities, they become the following product
formulas, which express given products of sines and cosines as
sums or differences :
(7-23) sin a cos ft = ^ [sin (a + 0) + sin (a - ft],
z
(7-24) cos a sin /3 = I [sin (a + 0) - sin (a - 0)],
(7-25) cos a cos = \ [cos (a + 0) + cos (a - ft)],
4
(7-26) sin a sin /3 = - [cos (a + 0) - cos (a - j8)].
1 44 Trigonometric Functions of Sums & Differences Sec. 7-4
To obtain the sum formulas, which express given sums or differ-
ences of sines and cosines as products, we first let
a + ]8 = x and a j8 = y.
Then, solving for a and /?, we have
= *-* and /3=^'
Substituting these values of a and /3 in (7-19), (7-20), (7-21),
and (7-22), we obtain the sum formulas. These are
(7-27) sin x + sin y = 2 sin -y cos
(7-28) sin a - sin y = 2 cos (^-y^) sin
(7-29) cos x + cosy =2 cos -^ cos
(7-30) cos x - cos = - 2 sin sin
Example 7-9. Express sin 3 a cos 5 a as a sum of sines.
Solution: Using (7-24) and replacing a by 5 a and by 3 a, we obtain
cos 5a sin 3a = ~[sin (5a + 3a) - sin (5a - 3a)] = ^[sin 8a - sin 2a],
2 -^
Example 7-10. Express cos 40 + cos 26 as a product of cosines.
Solution: Using (7-29) and replacing x by 40 and ?/ by 20, we have
A n I o/j ^fl _ OA
cos 4.6 -f cos 20 = 2 cos - r cos r = 2 cos 30 cos 0.
Z 2i
n 11 T^ xi i x-x s i n ^ sin 5x ,
Example 7-11. Prove the identity - = ; -- =- = tan x.
r cos 7x + cos 5z
rt , A . sin 7x sin
Solution:
rt /7a? + 5z\ . /7
2 COS ( ) Bin (
\ 2 / \
cos 7x -I- cos ft
COS
2 cos 6x sin x _
; __ _ tan x
2 cos 6# cos x
EXERCISE 7-3
In each of the problems from 1 to 10, write the given expression as a sum or
difference of two sines or two cosines.
1. sin 30 cos 40. 2. 2 sin 40 cos 20. 3. 2 sin 60 cos 40. 4. sin cos 40.
5. cos 40 cos 20. 6. 2 sin 65 cos 15. 7. sin 28 sin 20. 8. cos 21 cos 31.
9. sin 50 sin 0. 10. sin 110 sin 30.
Sec. 7-4
Trigonometric Functions of Sums & Differences
145
In each of the problems from 11 to 20, write the given expression as a product of
sines and cosines. Hint: In problems 18, 19, and 20, note that cos 6 = sin (90 - 6).
11. sin 30 + sin 20.
13. sin 60 + sin 30.
15. cos 80 - cos 20.
17. sin 40 + sin 25.
19. sin 40 + cos 44.
Prove each of the following identities:
21. sin + cos = V2 cos (^ ~ T) '
cos , sin
23.
25.
27.
A* -r . = COB 30.
sec 40 esc 40
sin 20 4- sin 40 p
cos 20 -f cos40~ tan ^'
sin a + sin ft __ a -f
12. cos - cos 40.
14. sin 40 + sin 20.
16. sin 30 - sin 80.
18. sin 64 + cos 38.
20. sin 65 - cos 33.
oo j. , L a COS ( ft)
22. tan a -f cot ft = ~~ *
cos a gin p
9A s * n _L cos
&* e\/\ i rTTJ ;
26.
sin 29
cos 30 + cos
sin n
7; = tan 0.
= tan
cos a -f cos ft v< *" 2
28. sin (0 + -j) + sin (0 - -|) = V2 sin 0.
29. sin (0 + -| ) - sin (0 - -|) = \/3 cos 0.
30. cos (-| + 0) - cos (-| - 0) = - V3 sin 0.
31.
sin cos
= tan -^
sin + cos + 1 2
32. sin (a + ft) sin (a - ft) = sin 2 a - sin 2 j8 = cos 2 ft - cos 2 a.
33. cos (a + ft) cos (a - /3) = cos 2 a - sin 2 j8 = cos 2 ft - sin 2 a.
Graphs of Trigonometric
O Functions; Inverse
Functions and Their Graphs
(-1,0)
8-1. VARIATION OF THE TRIGONOMETRIC FUNCTIONS
In Section 3-2, the trigonometric functions were defined in terms
of the coordinates (x, y) of the
point P(t), w r here the number t
represents the directed length of
the arc of a unit circle measured
from the point (1,0). Later, in
Section 3-9, an equivalent defini-
tion was given in terms of an
angle 9 in standard position. We
shall now consider the variation
of the trigonometric functions in
the four quadrants as f, and with
it 9, increases from to 2n. From
Fig. 8-1, we can read off the vari-
ations shown in Table 8-1.
For example, by noticing the changes in y as t increases con-
tinuously from to 27T, we find that sin t varies from to 1 in the
first quadrant, from 1 to in the second, from to 1 in the third,
and from -1 to in the fourth. Similar considerations lead to the
results for the cosine and tangent.
Recalling that esc t - - > we know that if either of these f unc-
sin t
tions increases the other decreases. Hence, the variation in esc t
can be determined from the variation in sin t. Similarly, we may
learn about the variation of sec t from that of cos t, and about the
variation of cot t from that of tan t.
We found in Example 3-2 that tan ?r/2 is undefined, which means
that tan t has no value when t = 7r/2. For the sake of easier tabu-
146
FIG. 8-1.
See. 8-2
Trigonometric Functions; Inverse Functions
147
TABLE 8-1
Variation of Trigonometric Functions
From To
From To
From To
From To
t
7T/2
7T/2 7T
7T 37T/2
37T/2 27T
sin i
1
1
-1
-1
cos t
1
-1
-1
1
tan t
oo
oo
oo
oo
CSC t
c 1
1 oo
-1
-1
sec t
1
-1
-1 -oo
1
cot t
oo
oo
oo
lation of this result in Table 8-1 we have employed the much
used symbols oo (infinity) and -co. These symbols merely signify
that in the neighborhood of Tr/2 or one of its odd multiples the
value of tan t is very large numerically. They are not to be used as
numbers.
8-2. THE GRAPH OF THE SINE FUNCTION
To construct a graph representing the variation of the sine, we
let x denote a real number or the value of an angle measured either
in radians or in degrees, and we let y denote the corresponding
value of the function. Corresponding values of x and y are plotted
FIG. 8-2.
148
Trigonometric Functions; Inverse Functions
Sec. 8-2
as points on a rectangular coordinate system. We can infer the
general appearance of the curve from the results summarized in
the preceding section, but an exact representation is more readily
obtained by using a table of sines.
Let us now construct the graph ofy = sin x from x = 7r/2 to
x = 2?r. The following values, found by the methods of Section 3-2,
are used to obtain the curve in Fig. 8-2.
X
7T
~ 2"
7T
~ 3~
7T
"IF
7T
6"
7T
T
7T
2"
27T
3
57T
6
y
-i
-0.87
-0.5
0.5
0.87
1
0.87
0.5
x
7T
77T
6
47T
3
3?r
2
STT
3
UTT
6
27T
y
-0.5
-0.87
-1
-0.87
-0.5
While the choice of a scale is arbitrary, a better propor-
tioned graph results if the same unit of length is used on both
axes. The unit so selected will represent the number 1 on the y-axis
and one radian on the #-axis. In terms of this unit a suitable length
can then be marked off on the x-axis to represent 2-rr or 360.
8-3. THE GRAPHS OF THE COSINE AND TANGENT FUNCTIONS
Using the table of trigonometric functions, the student should
make a table of corresponding values for y = cos x and one for
FIG. S-3.
Sec. 84
Trigonometric Functions; Inverse Functions
149
7T/2
37T/2
27T
**tzn x
FIG. 8-4.
y = tan #, similar to that used for y = sin # in Section 8-2. Study
the graphs in Fig. 8-3 and Fig. 8-4 on the basis of the tables you
have made.
If we compare Fig. 8-3 with Fig. 8-2, we see that the graph of
y = cos x may be obtained from the graph of y = sin x by moving
the graph ofy = sin x to the left a distance of rr/2 units. This fact
can be checked by using the relationship cos x = sin (# + ir/2),
from which it follows that cos = sin 77/2, cos vr/G = sin 27T/3, and
so on.
8-4. PERIODICITY, AMPLITUDE, AND PHASE
The trigonometric functions are among the simplest of a large
class of functions which are periodic. As a preliminary to defining
periodic functions, we shall call attention to some examples of
phenomena which recur periodically, such as the rotation of the
earth about its axis, sound and water waves, the vibration of a
spring, and many other vibratory and wavelike phenomena. The
behavior of the object involved in a phenomenon of periodic nature
determines the type of function that is required to represent it
properly. We note particularly that, because of the recurrence
characteristic of such a phenomenon, the values which the function
assumes in any interval of given length are also taken on in any
other interval of the same length. This statement apparently indi-
cates that a function of x is periodic with period p if, for every
value of x, the function returns to the same value when x is
increased by p. More specifically, we state the following.
1 50 Trigonometric Functions; Inverse Functions Sec. 84
Definition. A function f(x) is said to be periodic if there is a non-
zero number p for which
f(x + p) =/(x)
for all numbers x in the domain of f(x). Any such number p is
called a period; the smallest positive number p satisfying the
requirement is called the period.
Evidently, if p is the period of f(x), then np is a period, for
every integer n.
Periodicity of Trigonometric Functions. No matter which of the
two viewpoints is considered in the definition of the trigonometric
functions, we shall see that, if g t ^ 2ir, then
any trigonometric function of (t + 2?r) = same function of t.
According to the definitions given in Sections 3-1 and 3-2, P(t)
and P(t + 27r) represent the same point on the circumference of
the unit circle. To locate P(t) we start at (1,0) and proceed
around the circle in the proper direction a distance of \t\ units. To
locate P(t + 27r) we continue another 2?r units from the point P(t) .
This merely adds another complete revolution, and we arrive at the
same point P(t).
If we consider the definitions of the functions in terms of angles,
as given in Section 3-9, we note that the angle + 2n is coterminal
with and that any trigonometric function has the same value for
coterminal angles.
Period of Sine, Cosine, Cosecant, and Secant. From a study of the
sine curve, it is apparent that sin x assumes all values between 1
and +1 as x, starting from any value, varies through 2ir units. In
other words, the graph of the function repeats itself during each
interval of length 27r, for positive and negative values of x. Or
stated more concisely,
sin (x + 2?r) = sin x.
This is equivalent to saying that sin x is a periodic function of x,
and that 2?r is a period. It remains now to show that 2ir or 360 is
the smallest positive number p for which sin (x + p) = sin x and
for which cos (x + p) = cos x.
Since, by definition of a period p, sin (x + p) = sin x for any value
x 9 we shall select for the purpose of our proof the particular value
x = ir/2. We then have
. /7T * \ . 7T 1
sin (- + P) = sm 2" = L
But since sin f -5- + pj = cos p, it follows that cos p = 1.
Sec. 8-4
Trigonometric Functions; Inverse Functions
151
Hence, p must be an even multiple of TT. The smallest even multiple
of TT is 27r, which must also be the smallest positive period of the
sine function.
In a similar manner, we find that 2rr is also the period of the
cosine function. Because of the reciprocal relationships existing
between the sine and cosecant and between the cosine and secant,
27r is also the period of the cosecant and the secant.
Period of Tangent and Cotangent. To find the period of the
tangent, we write
tan (x + p) = tan x }
and we let x = 0. Then tan p = 0, and we find that TT is the period
of the tangent. A similar argument shows that TT is also the period
of the cotangent.
Period of sin bx. We have just seen that the period of sin x is
27T. We shall now determine the period of sin bx, where 6 is a
positive constant. That is, we want to know the smallest positive
change in x which will produce a change of 27r in bx. If p repre-
sents this change in x, we can find p from the relationship
b(x + p) = bx + 27T.
Solving for p, we immediately find that
27T
P = T -
Thus, the period of sin bx is equal to the period of sin x divided by
b, that is, 2;r/&.
Similarly, it can be shown that 27T/6 is also the period of cos bx,
esc bx, and sec bx. It is also true that the period of tan bx and
cot bx is 7r/b.
FIG. 8-5.
1 52 Trigonometric Functions; Inverse Functions Sec. 8-4
Let us consider the graph ofy = sin 2x shown in Fig, 8-5. Since
the period of sin 2x is 2ir/b = 2rr/2 TT, the function will assume
the same range of values in the interval from to TT that sin x takes
in the interval from to 27r.
Note that the graph of y = 2 sin x is similar in form to that of
y = sin x, which is shown in Fig. 8-2. The period is 2ir for both
curves. However, for any given value of x, the corresponding value
of y in y = 2 sin x is twice as large as is the corresponding value of
y in y = sin x for the same value of x. In the graph ofy = a sin x,
where a > 0, the greatest value of y is a, and the smallest value of y is
a. The constant a is called the amplitude of the function or of the
graph. Thus, the amplitude of the graph of y = sin x is 1, while that
of the graph of y = 2 sin x is 2.
In general, for the function y a sin x, where a is a real number,
the amplitude is equal to |a| and the period is equal to 27r. Also, in
general, for the graph of y = a sin bx, where a and b are real, the
amplitude is |a| and the period is
Phase Angle. Since the graph of y = cos x may be obtained from
the graph ofy = sin x by shifting it to the left a distance equal to
7T/2 units, we say that the graph ofy = cos x differs in phase by
77/2 from the graph ofy sin x. The amount of horizontal dis-
placement of two congruent graphs, amounting to Tr/2 radians in
this case, is called the phase difference, or the phase angle. The
amplitude and the period are the same for y = cos x as for y = sin x.
y cos 2x - sin (2x + 7T/2) = sin 2 (x + 7T/4)
FIG. 8-6.
Now consider the graph of y = cos 2x - sin 2(x + ~) in Fig.
8-6. This curve may evidently be obtained from that ofy = sin 2x
in Fig. 8-5 by a shift of 7r/4 units to the left in the x direction.
Hence, the graph of y = sin (2x + ~) differs in phase from that
4U
ofy = sin 2x by 7r/4 radians.
Sec. 84 Tr/gonomefric Functions; Inverse Functions 153
A simple method for finding the phase displacement is to locate
the point near the origin for which the function sin (2x+^)
equals zero ; that is, to find the smallest numerical value of x that
makes the quantity x + j zero. We have then sin2(#-f ~) =0
when x + - = or when x = 7r/4. Hence, the phase difference is
7T/4 radians, and the shift of the graph is toward the left since the
sign of x is negative.
It can be shown that, in general, the phase displacement of any
trigonometric function of (bx + c), where b > 0, is to the right or
left by \c/b\ radians (or degrees). The direction of the displace-
ment depends on whether c/b is negative or positive.
Finally, we arrive at the conclusion that for the graph of y =
a sin (bx + c) the amplitude is \a\ and the period is 27T/6. Also, its
phase differs from that ofy = sin x by c/b.
Because of its usefulness in many applications, we shall illustrate
by means of an example a procedure for reducing an expression
of the form A sin + B cos & to the form a sin (9 + a) .
Example 8-1. Reduce A sin + B cos to the form a sin (0 + a).
Solution: Write
A sin + B cos = \/A 2 + B 2 (~ _ sin + = cos 0^
\\/A 2 + B* VA 2 + B 2 /
The absolute values of the coefficients and = cannot be greater
' An -f B 2 \/A 2 -f B 2
than 1, and the sum of their squares is 1. Hence, they may be taken as the cosine
and sine, respectively, of some angle a. Therefore, the expression A sin -f B cos B
becomes
+ B 2 (cos a sin 6 + sin a cos 0) = \M 2 4- B 2 sin (6 + a),
where A = a cos a and J5 = a sin a.
Example 8-2. Express 4y = sin 26 - V3 cos 26 in the form y = a sin (60 + a),
and sketch the graph.
Solution; Since A = 1 and B = \/3, we have \/^ 2 -h # 2 = 2. Therefore, $
sin 26 -f- \/3 cos 20 = 2^ sin 20 - ^ cos 20Y
If we identify this result with the expression \/A* -f B 2 (cos a sin 20 + sin a cos 20),
we have cos a = ^ and sin a = ~~ It follows that a is a fourth quadrant
angle and may be taken equal to ir/3.
154
Trigonometric Functions; Inverse Functions
Sec. 8-4
We have, finally, - _ - .
y = i(sin 26 - V3 cos 29) = | sin ^20 - y J
The amplitude of this function is r > its period is TT, and its phase angle is ir/6.
4
i
1/2-
T
S~
\ , /
^N /",
-1/2-!
A/6 7T/2 \ 7T /
r -l/2sin(20-7T/3)
37T/2 \ 27T /
FIG. 8-7.
Since the interval from (TT/G, 0) to (77T/6, 0) is one period in length, the part of the
curve obtained for this interval may be repeated indefinitely in both directions to
give the complete curve. The curve is shown in Fig. 8-7. (Here for convenience
we have employed different units of length on the two axes.)
EXERCISE 8-1
In each of the problems from 1 to 24, find the period, amplitude, and phase angle
of the trigonometric function.
1. 3 sin 0.
4. cos =
2.sin|.
5. = sin 7 6
3 4
7. 4 tan \ 8
o
8. si
o
11. 3 sin 50.
14. sec 90.
17. tan (TT + 4).
i
19. esc (ir6 + 7). 20. cot (2?r - TT).
22. 6 - cos 40. 23. 5 + 3 sin (26 - -
\ o ,
10. sin 7T0.
13. cot 60.
16. 3 esc 37r
3. ^ cos 0.
6. 2 cot |
9. cos0-
4
12. 5 tan 7T0.
15. 5 cot 40.
18. cos (30 - 2).
21. 2 + sin 0.
24. 4 + 2 cos (20 + -|
\ o ,
In each of the problems from 25 to 30, sketch the graph of the given function
by constructing a table of values.
25. y = cos x
6
28. y = 5 cos \ x
26. y = tan ^ z
29. y = 2 tan |
27. y = 2 sin 3.r.
30. y = 3 sin 2x.
Sec. 85 Trigonometric Functions; Inverse Functions 1 55
Sketch each of the following graphs without constructing a table of values.
2 1
31. y = sin ^ a:. 32. y = cos 4z. 33. # = tan x.
6 &
34. y = 2 cos 3.r. 35. ?/ = 3 cos irx. 36. */ = r tan 3x.
4U
37. y = I sin 1 3. 38. y = | cos | x. 39. ?/ = 5 cos (x + ~
40. !/ = cos fa? + 2). 41. y = sin (x - ^) * 42 * V = cos ( 3x ~ 2 )-
43. ?/ = sin (2x + 1). 44. y = cos (2wx - TT). 45. y = sin f j x + 7 V
46. ?/ = sin + cos 0. 47. ?/ sin 6 - cos 0.
13
48. y = \/3 sin 20 + V5 cos 20. 49. y = cos ^ ~ 7 sin ^
50. y = V2 cos 30-3 sin 30. 51. y = 2 sin - cos 20.
8-5. INVERSE FUNCTIONS
In Section 2-3, we defined a function by setting up a rule of
correspondence between twa sets of numbers, X and 7, called the
domain of definition of the function and the range set of the func-
tion, respectively. The function was called single-valued if just
one number y of the set Y is assigned to each number x of the set X.
If more than one number of Y corresponds to some value of x, the
function is multiple-valued.
If we know that y f(x), we may pose a reverse problem. We
assume y to be given and ask for all corresponding values of x.
Naturally, y is limited to lie in the range of the given function,
since otherwise no x exists. The function which makes correspond
to each such y all values of x for which y = f(x) is called the
inverse function corresponding to the given function.
We shall begin our discussion of inverse functions with an
example in which X is the set of all real numbers, Y is the set of
all non-negative real numbers, and the correspondence is deter-
mined by the relationship
y = x 2 .
Ordinarily, we assign values to x in order to calculate values erf
x 2 . In this case, we have a rule of correspondence that assigns just
one number y to ea$h chosen number x. Hence, y is a single-valued
function of x. The s graph ofy = x 2 is shown in Fig. 8-8.
Assume now that y is given and that we wish to determine cor-
r^sponding values of x. To do this we solve the given equation
x 2 = y for x, and obtain two numbers x = \/y and* x = ^/y cor-
responding to every, non-negative number y. Since the rule of
156
Trigonometric Functions; Inverse Functions
Sec. 8-5
correspondence assigns two values of x to each chosen number y,
we see that # is a double-valued function of y. In this case, the
admissible values of y are restricted to zero and the positive real
numbers, while those of x comprise all real numbers, as was indi-
cated in the specification for the sets X and Y.
O
FIG. 8-8.
FIG. 8-9.
In the study of mathematics, we generally prefer to use the sym-
bol x to represent the independent variable and y to represent the
dependent variable. To be consistent with this preference, we shall
call y = x 2 and y = .^/x inverse functions, each being called the
inverse of the other.
The graph ofy = \/Hc is obtained by plotting that of x = y 2 , as
shown in Fig. 8-9. We note that the roles of x and y are interchanged
in the two equations y = x 2 and x = y 2 . Thus, we see that the
curve in Fig. 8-9 is actually the curve of Fig. 8-8 with the axes
interchanged and one of them reversed in direction.
8-6. INVERSES OF THE TRIGONOMETRIC FUNCTIONS
The Inverse Sine. Let us consider the function y = sin x and
attempt to apply a discussion similar to that in Section 8-5. Here,
as has been noted, the domain is the set of all real numbers, while
the range is the interval 1^7/^1. Referring to the graph of
y = sin x in Fig. 8-2, and recalling the periodic properties of this
graph, we see that, for every given number y such that 1 :f y 2S 1,
there are infinitely many values of x such that y = sin x. To desig-
nate the totality of all values of x such that y = sin x, we write
x = sin"" 1 y,
which is read x is the inverse sine of y. The student should note care-
fully that the symbol sin" 1 y must be distinguished from (sin y)~ l 9
which equals - or esc y.
sin t/
Sec. 8*6 Trigonometric Functions; Inverse Functions 1 57
Another notation that is frequently used to represent this inverse
function is
x = arc sin y,
which is read x is the arc sine of y.
In order to conform to the preferred practice of considering y
as a function of x, we may designate the inverse of the sine func-
tion by writing
y = sin" 1 x or y = arc sin x.
We note the following properties of this inverse function.
If # is a number such that \x\ > 1, then y does not exist. This
property follows from the fact that the sine function takes on only
the values from -1 to 1. Hence, the inverse sine function sin* 1 x is
defined only when 1 ^ x =i 1. For example, sin' 1 2 is not defined,
since there is no number or angle whose sine is 2.
If # is a number such that \x\ ^ 1, then y certainly exists. More-
over, because of the periodicity of the sine function, there are
infinitely many values of y = sin * x corresponding to every such
value of x. For example, it x = 1/2, then y = sin' 1 1/2 means that
y is any number or angle such that sin y 1/2. Then y may be
taken as ?r/6, 5-77/6, or any value that differs from these by integral
multiples of 27r. The totality of these values of sin- 1 1/2 may be
represented as
+ 2n?r and + 2mr, where n = 0, 1, 2, .
We shall find it convenient to plot the graph of y = sin' 1 x for a
further study of the inverse function. Since y = sin' 1 x and x = sin y
express exactly the same relation between x and y, the graph of
Fig. 8-2 may be used as a graph of the inverse function. We obtain
the graph ofy = sin* 1 x simply by interchanging the axes in Fig.
8-2 and reversing one of them in direction. The result is shown in
Fig. 8-10.
The question now arises whether the t/-axis can be subdivided
into intervals within each of which y has just one value correspond-
ing to each x such that \x\ ^ 1. One way of doing this is by select-
ing a first interval
TT ^ ^ TT
-2 ***2"
Other intervals, such as -^ ^ y ^ - and ~ O w g y ^ ~ > are then
z z z z
selected. With the entire y-axis thus subdivided, we may think of
the graph ofy = sin" 1 x as consisting of the graphs of infinitely
many single-valued functions or branches.
158
Trigonometric Functions; Inverse Functions
Sec. 8-6
7T/2
-i
3TT/2
O i
-7T/2
FIG. 8-10.
-1
37T/2
O /I
"-7T/2
FIG. 8-11.
To avoid any ambiguity in later applications as a result of this
multiple-valued property of the inverse sine, we shall often restrict
y so as to make the function single-valued. There will then be but one
value of y corresponding to each value of x such that sin y = x. We
shall determine this value from the branch for which -- ^ y ^ -
This branch is called the principal branch. The values of y chosen
from this branch are called the principal values of the inverse-sine
function and are represented by the equation
y = Sin" 1 x or y = Arc sin x.
Note that in this case the initial letter of the name is capitalized.
This restriction to the two quadrants containing the smallest
numerical values of y results in a single-valued function y = Sin- 1 x.
The values of y are such that - ^ y ^ f or x in the interval
-1 g x ^ 1. For example, we have Sin- 1 (-1) = -7r/2, Sin- 1 (-1/2) =
-7T/6, Sin- 1 = 0, and Sin- 1 1 = ir/2.
The Inverse Cosine. We shall next consider the function y = cos x.
With the help of Fig. 8-3, we see that the range of y = cos x is the
interval 1 S y ^ 1, and that for every y in this interval there are
infinitely many values of x such that y = cos x. We are thus led to
the inverse cosine function, which makes correspond to y all the
values of x such that y = cos x. If again we interchange the sym-
bols x and y, we may write the inverse cosine function as
y = cos~ x x or y = arc cos x.
Sec. 8-6
Tr/0onomefr/c Function^; Inverse Functions
159
This inverse function has the following properties.
If # is a number such that \x\ > 1, then y does not exist, because
there is no value of y for which |cos y\> 1.
If x is a number such that \x\ ^ 1, then there are infinitely many
values of y designated by cos' 1 x.
The graph ofy = cos* 1 x is shown in Fig. 8-11. The method for
plotting it is similar to that used for graphing y = sin' 1 x. We first
write x cos y. We then plot a cosine curve by proceeding as for
Fig. 8-3, except that values of the independent variable y are laid
off on the ?/-axis.
The principal branch of the curve in Fig. 8-11 is the portion of
the curve for which g y ^ TT. It is represented by the principal
value of the function, which is denoted by
y = Cos" 1 # or y = Arc cos x.
The Inverse Tangent. The inverse tangent function is
y = tan" 1 x or y = arc tan x.
-3 -2 -1
7T/2
o
-7T/2
-7T
FIG. 8-12.
It is represented by the graph in Fig. 8-12. We note that there are
infinitely many valu.es of y for every value of x.
The principal branch of the graph of y = tan- 1 x is the portion
for which ~ < y < ~ This is represented by the equation
y = Tan" 1 x or y = Arc tan
160
Trigonometric Functions; Inverse Functions
Sec. 8-6
The Inverse Cotangent, Secant, and Cosecant. The other inverse
trigonometric functions are :
y = cot" 1 x or y = arc cot x,
or y = arc sec x,
or y = arc esc #.
\Y
y = sec" 1 a:
?/ = esc" 1 a:
FIG. 8-13.
We shall show only the graph ofy = cot- 1 x. See Fig. 8-13. The
principal branch of this curve is given by < y < TT.
Principal Values of the Inverse Cosecant and Secant. The selection
of principal values of ?/ = esc- 1 x and y = sec- 1 x is by no means uni-
form among all authors. Some writers adopt the range between
and 7T/2 for both functions when x is positive, and between TT and
7T/2 when x is negative. The authors prefer, however, to use the
definitions
Csc- 1 x = Sin- 1
and
Sec" 1 x = Cos" 1 (-)
We have, therefore, the following ranges of principal values of
the inverse trigonometric functions :
7T 7T
-o- SI Sin" 1 x ^ > ^ Cos" 1 # ^ TT,
< Cor 1 a; < TT,
- ^ Csc-
1 * ^ -J (Csc^ 1 x * 0), ^ Sec" 1 x ^ TT (Sec" 1 x ^ y)
Sec. 8-6 Trigonometric Functions; Inverse Functions 161
(Ox
Sin- 1 =
o
3 34
Solution: Let Sin- 1 -= = 6. Then sin = ^ > and cos = = Since only the
5 5 5
principal value of sin- 1 = is used, is a first-quadrant angle and cos cannot equal
o
4 / 3\ 4
= Hence, cos (Sin- 1 - J = -
Example 8-4. Find the value of sin [Tan- 1 ( - x)],
x being a positive number.
Solution: Let Tan' 1 ( - x) = 0. Then tan 6
= x, and lies between w/2 and 0. If angle
6 is constructed in standard position, as shown
/
in Fig. 8-14, then sin 6 is found to be
Hence, sin [Tan" 1 ( - x)] =
FIG. 8-14.
Y
1
O
*l
-
Example 8-5. Find Arc cos (cot 60).
Solution: Let Arc cos (cot 60) = 6. Then cos 6 = cot 60 = .5774, by Table II.
Therefore, the value of 6 is Arc cos .5774. Since 6 is restricted to the interval from
to 180, 6 must, in this case, be a first-quadrant angle. Hence, = Arc cos .5774
= 5444', and Arc cos (cot 60) = 5444'.
EXERCISE 8-2
In each of the problems from 1 to 6, find the inverse of the given function.
O/y. K O-r 1 1
< - ^- 5 - _/*-! 3 y=rnx+b.
_
3
5. y =
6
2x - 1
+6
3
2'
3 4
In each of the problems from 7 to 22, find the value of the given expression.
8. Tan- 1 1.
9. Arc cos
11. Cot- 1 0. 12. Sin- 1 ( - J) 13. Cot- 1 ^
V &/ o
10. Cos- 1 (- 1).
14. Csc-i 1.
15. Tan-'
16. Cos- 1
19. Cos- 1 ( - ^
21. Sin- 1 ^-0.414).
17. Tan- 1 ( - V). 18. Arc tan (- 1).
20. Tan- 1 [sin (- ir/2)].
22. Tan- 1 (-1.414).
In each of the problems from 23 to 37, evaluate the given expression.
23. tan (sin- 1 j|) 24. sin (sin- 1 1) 25. sin
26. cot Sin- 1
27. cos
(Cos' 1 1)
281 tan (Sin- 1 .6450).
162
29. sin (Cos- 1 .9200).
32. sin (Cot- 1 7
V 4
35. cc
Trigonometric Functions; Inverse Functions Sec. 8-6
31. sec ( Sin- 1 i)
\ 5/
34. sin (Cot- 1 1)
30. tan (Cot- 1 2).
33. cot (Cos- 1 1)
36. cos f Sin- 1 5
\ o
37. Sin- 1 (tan ^
In each of the problems from 38 to 56 simplify the given expression, taking u as
a positive number. In 38 to 40, 54, 56, u O/2. In 44, 46, 48, 55, u <1.
38. Cos- 1 (sin u). 39. Sin- 1 (- sin u).
40. Sin- 1 (cos u).
42. sin (Sin- 1 u).
41. esc fsin- 1 i
\ u
43. tan (Tan- 1 u).
45. cot ("Tan- 1 - _
V VI +
U
u
48. esc (Sin-
47. tan ( Sin- 1
\
49. tan (Cos~ l
V v
51. sec (Cos- 1 u).
53. sec (Sin- 1 14).
55. Cos- 1 (cos
+ u
50. cos (Tan- 1 u).
52. cot (Sin- 1 u).
54. Sin" 1 (sin u).
56. Cot- 1 (tan u \
In each of the problems from 57 to 65, draw the graph of the given function by
changing from the inverse function to the direct function and using the period,
amplitude, and phase angle of the function to assist in plotting.
57. y = --sin- 1 x.
Q
59. y = cos- 1 a:.
61. y = - cos- 1 x + 2.
o
+ 1).
58. y = 2 tan- 1 x.
60. y = ^ sin- 1 sc - 1.
Zi
62. ?/ = 4 tan- 1 z 4- 3.
64. y = 3 sin- 1 (2x + 1) - 2.
63. y = (Cos- 1
65. y = 4 tan- 1
66. Prove that sin (Cos- 1 u) ^ 0, if < u g 1.
67. Prove that cos (Sin- 1 it) 0, if < u 1.
68. Prove that cos (Tan- 1 u) > 0, if u ^ 0.
69. Prove that tan (Cos- 1 it) 0, if < u 5 1.
70. Prove that Sin- 1 u + Sin- 1 (- ti) = 0, if - 1 u 1.
71. Prove that Tan" 1 w + Tan- 1 ( - u) = for all values of u.
72. Prove that Tan- 1 u + Tan- 1 (l/i*) = y > if w > 0.
73. Prove that Cos- 1 u 4- Cos- 1 (-u)=ir, if-lgul.
74. Prove that Tan- 1 tt - Tan~* v = Tan- 1 j** " J 'if ti > and v > 0.
V linear Equations and Graphs
9-1. SOLUTIONS OF SIMULTANEOUS EQUATIONS
Often problems arise that involve two or more unknowns and as
many equations. The solution of such a problem requires the deter-
mination of numbers which simultaneously satisfy the given equa-
tions. Equations for which we seek common solutions are referred
to as a system of simultaneous equations. If the system has at least
one solution, it is said to be consistent; otherwise, it is called
inconsistent.
Let us investigate the following pair of simultaneous linear
equations :
f aix + biy = a,
(9-1)
( a 2 x + b 2 y = c 2 .
It is desired to find all pairs of values of x and y which satisfy both
equations, excluding from consideration the cases a\ = &i = and
a 2 = & 2 = 0.
We proceed by multiplying both sides of the first equation by & 2
and both sides of the second by bi, obtaining
b 2 biy = c 2 bi.
Subtracting the second equation of this pair from the first, we
obtain
(9-2) (aib 2 a 2 bi
If (aib 2 02&i) ^ 0> i we ^ n d that
_
Thus, we see that we have exactly one value for x in any solution
which may exist.
163
1 64 Linear Equations and Graphs Sec. 91
By multiplying the original equations by a? and a lf respectively,
^nd subtracting the first equation thus obtained from the second
one, we obtain
(9-3) (ai& 2 - a 2 bi)y = aic 2 - a 2 a.
If (ai&2 - a 2 6i) T* 0, we find that
d\C2 CL 2 C\
4i ~~ _____________
a\b 2 CL 2 bl
Again, we have exactly one value for y in any solution which may
exist.
Hence, if (di& 2 &2&i) is not zero, we have at most one solution
of the pair of given equations, namely,
(9-4) ' x = l ~ l > y -
Substitution of these values for x and y in (9-1) shows that we
really have a solution. The reader should verify this solution by mak-
ing the substitutions.
Consistent and Independent Equations. If (eii& 2 e^&i) ^ 0, we
have exactly one solution of (9-1), which is expressed by (9-4),
and the given equations are called consistent and independent. We
may consider, as an example, the pair of equations
I 2x + 3y = 24,
\ 5x-2y= 22.
Here a^b 2 - 0361 = (2) (-2) - (5) (3) = -19, and the values x = 6
and y = 4 give a solution. In other words, when these values are
substituted in the two given equations, both equations are satisfied.
Thus,
I 2(6) + 3(4) = 24,
I 5(6) - 2(4) = 22.
Cases with ai& 2 - 0261 = 0. Let us now consider cases in which
di&2 &2&1 = 0. If di ^ 0, division by di gives
, 2 ,
02 = 01.
01
We then define k = > and we have
ai
(9-5) a 2 = fan, 62 = fc&i.
Now let ai = 0. Then, since 61 cannot be zero by our initial assump-
tion, and since a^bi = 0, we have a 2 = 0. In this case, if we define
k = -2. , we see that (9-5) again follows. Thus, ai& 2 - a 2 bi = has
Sec. 9-1 Linear Equations and Graphs 1 65
been shown to mean that the left members of (9-1) are
proportional.
We note also that if ai& 2 0261 = 0, equations (9-2) and (9-3)
become
f X = Ci&2 C2&1,
(9-6)
( y = aic 2 - a 2 ci.
It is clear that these equations cannot be satisfied by any pair of
values of x and y, unless both right members are also zero. Accord-
ingly, if we wish to determine whether or not the given system has
a solution when ai& 2 &2&i = 0, it becomes necessary to take into
account the two cases of the right sides of (9-6) being zero or not
zero.
Consistent and Dependent Equations. In the first case referred
to in the preceding paragraph, a,ib 2 o^bi = 0, and Ci& 2 ^2&i and
a,iC 2 a 2 Ci are also equal to zero. Hence, substituting the values
of & 2 and 02 from (9-5), we have
(ci/b 2)61 = (ci/fc c 2 )ai = 0.
Since a lf bi are not both zero, it follows that c 2 = kci. This means
that one of the original equations is a constant multiple of the
other, and any pair of values of x and y that satisfy one equation
will also satisfy the other. Under this condition, the equations are
said to be consistent and dependent.
We have as an illustrative example the equations
f 2x + 3y = 24,
( 4z + Qy = 48.
Here ai& 2 - a 2 b l = (2) (6) - (4) (3) = 0. Also, the coefficients of
the unknowns and the constant term in the second equation are
multiples of the coefficients of the unknowns and the constant term
in the first, and the multiplier is 2 for all three terms. Infinitely
many solutions exist. Some of them are : x = 0, y = 8 ; x = 12, y = ;
04 2t
and x = t,y - , where t is any number.
o %
Inconsistent Equations. Finally, let us suppose that in the orig-
inal equations aj)* a^bi = and at least one of the numbers
Ci& 2 ~ c 2 6i, aiC 2 a 2 Ci is different from zero. Hence, no solution of
(9-1) exists and the equations are inconsistent. This case is char-
acterized by the condition that one of the numbers (cik <%) 61,
(cik 02)0,1 is not zero, in view of (9-5). It follows that c 2 ^ kci*
1 66 Linear Equations and Graphs Sec. 91
Hence, the multiplier for the left members of (9-1) does not apply
jto the constant terms.
Consider, for example, the equations
I 2x + 3y = 24,
1 4x + Qy = 7.
Here the coefficients of the unknowns in the second equation are
multiples of those in the first equation, and the multiplier is 2 for
both terms; however, the multiplier for the constants is not the
same as that for the other terms. Hence, these equations are
inconsistent.
9-2. ALGEBRAIC SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS
To solve a consistent and independent system of two linear equa-
tions in two unknowns, we reduce the system to one equation in one
unknown by eliminating one of the unknowns. The following
example will illustrate two commonly used methods for eliminating
the unknowns.
Example 9-1. Solve the equations
2x+3y= 24,
5x - 2y = 22.
Solution: Since a\b 2 a 2 &i = (2) (-2) - (5) (3) = - 19, the equations are
consistent and independent, and there is but one solution. If we use the method of
elimination by addition or subtraction, the procedure is the same as that indicated
in obtaining the solution (9-4) from equations (9-1).
To eliminate y t multiply the first equation by 2 and the second by 3, in order to
make the coefficients of y numerically equal in both equations. We thus obtain
4z + Qy = 48,
5x - 6y = 66.
Adding, we get 19$ = 114.
Solving for x, we have x = 6.
Now, substitute 6 for x in the first of the original equations. Then
3y = 24 - 2x = 24 - 12 = 12,
or
y=4.
Alternate Solution: If we use the method of elimination by substitution, we begin
by solving the first equation for y in terms of x. We thus get
24 -2x
Sec. 9-3 Linear Equations and Graphs 1 67
24 2x
We then substitute ~ - for y in the second equation and obtain
Solving for x, we have
Ux - 2 (24 - 2s) = 66,
or
15x - 48 + 4s = 66.
Hence,
19z = 114,
and
x =6.
Substituting 6 for x, as before, we find that y = 4.
Example 9-2. A grocer has some coffee selling at 80 cents per pound and some
at 90 cents per pound. How much of each must he use to get a mixture of 100
pounds worth 86 cents per pound?
Solution: Let x = number of pounds of 80-cent coffee, and y = number of pounds
of 90-cent coffee.
Then
x + y = 100,
and
O.SOz + 0.90y = 0.86 (100).
Simplifying, we have
x + y = 100,
z + 9y = 860.
These equations have the single solution x = 40, y = 60.
9-3. LINEAR EQUATIONS IN THREE UNKNOWNS
In the solution of a system of three equations in three unknowns,
one method is to employ the following steps, which we do not justify
here :
1. Eliminate one of the unknowns from a pair of the equations;
then eliminate this same unknown from another pair of the
original equations.
2. Solve the resulting two equations for the two remaining
unknowns.
3. Substitute the values found in step 2 in any one of the orig-
inal equations to find the third unknown.
Example 9-3. Solve the system of equations
2x + 3y - 2=5,
x - 5y + 2z = 1,
3x + y - 42 = - 1.
168
Linear Equations and Graphs
Sec. 9-3
Solution: Eliminate z from the first and second equations by addition to obtain
5x + y = 11.
Now eliminate z from the second and third given equations by addition to obtain
5z - 90 = 1.
We then consider the equations
5x + y = 11,
Solving these equations for x and y by subtraction, we have x = 2 and y = 1.
Substitution of these values in the first given equation gives z = 2. Hence, the
solution of the given system is z = 2, y = 1, 2 = 2.
EXERCISE 9-1
In each of the problems from 1 to 30, solve the given system of equations.
Check all solutions.
1. (3x - 2y = 6,
I x - 30 = 4.
N
'1
10. 1
13.
16.
18.
20.
3z + 2y = 1, 5. J
\
/M __ O-y K 1
3z + 20 = 4, 8. f
2z - 30 = 3. \
x + 20 + 1 = 0, 11. f
x - 40 + 2 = 0. 1
_2 3 _7
3x - + 22 = 4,
x + 20 - 32 = 1,
2s - 30 + 2=2.
a; + 20 + 32 + 1 = 0,
x - 40 + 62 + 1 = 0,
2x + 60 + 72 + 2 = 0.
x + 20 - 32 + 1 = 0,
- x + 30 - 42 - 5 = 0,
2x + 60 - 4
2+3=0.
22.
3x
+
20
+
=
2,
5a?
+
y
+
3 =
o,
2x
30
42
+
5 =
0.
24.
3*
-
2/
+ 42
=
2,
4s
+
40
+ 42
=
5,
2#
-
#
+ 62
=
9.
3x - y = 7,
2x + y = 8.
2x + 30 + 1 = 0,
3x - 2/4-7=0.
x + 2y = 3,
2x + 30 = 1.
2x + 30 - 1 = 0,
3x + 0+3=0.
17.
19.
21.
23.
25.
3.
6.
9.
12.
15.
- 3z = 6,
x + 20 = 3.
2x + 2y - 3 = 0,
5x + 30 + 4 = 0.
3z + 0+7=0,
4z + 80 + 9 = 0.
2x + 60 - 7 = 0,
3z - Sy + 9 = 0.
2x + 40 - 50 = 3,
3z + - 7 = 2,
4x + 80 - 10z = 1.
s + 20 - 2=3,
- x + 40 + 2s = 1,
3z + - 30 = 2.
- 3x + 80 + 9s - 3 = 0,
2x + 30 + 2-4=0,
Bx - 20 - 22 - 4 = 0.
3z - + 52 = 0,
x - 42 = 2,
4z - 20 - 32 + 1 = 0.
3x - 40 + 22 = 3,
2x + y =1,
5x - 3y 4- 42 + 5 = 0.
2x + 30 - 62 + 2 = 0,
50 + 3 - 2 = 0,
1 3 A
a- V-2 f ~2 =0 '
Sec. 9-4
Linear Equations and Graphs
169
26.
28.
By + 2z = 4,
2z -20 +3* =3,
3s + 4s = 2.
(4
1 _2 1 = 1
x y z ~~3
27.
29. 13(3 +0) - 4(3 -0) =5,
1 4(3 -f 0) 3(3 - 0) = 5.
30. (3(3 -f 0) -4(3 -0) =5,
(X - 0) - 3(3 + 0) = 7.
(X I' Z *
For each of the following systems of equations, determine whether it is consistent
and independent, consistent and dependent, or inconsistent.
31. /23 - 30 = - 5, 32. (33 - 50 -f 8=0, 33. J33 - 20 = 8,
\ x + 80 - 10 = 0.
35. j 2x - 70 + 1 = 0,
[210 - 63 - 3 = 0.
38. f - 9z + 120 = 3,
3x + 40 = 1.
1 43 - 60 = - 3.
34. [2x + 40 = 3,
\ x + 20 = 6.
37. /23 - 30 = 1,
5 + 6w = 2.
40.
- x + 30 = 2,
2x = 60 + 14.
[63 - 40 = 8.
36. J45z - 270 = 21,
J153 - 90 = 7.
39. | 43 - 30 = 6,
[150 -203 +6 =0.
ay + 63 = a6.
bx -{-ay = a6c.
9-4. GRAPHS OF LINEAR FUNCTIONS
The discussion of rectangular coordinates in Section 2-1 set the
stage for the pictorial representation of a function. By this repre-
sentation of a function f(x), we mean the graph of the equation
y = f(x). It consists of all points, and only those points, whose
coordinates x and y satisfy the equation.
In the same section we considered the graphing of lines parallel
to the coordinate axes. The equation x = 3 was shown to represent
a vertical line, that is, a line parallel to the y-axis which intersects
the #-axis at the point (3, 0). This line thus includes all points 3
units to the right of the t/-axis. This example illustrates the fact
that a linear equation in x alone represents a line that is parallel
to the ^/-axis. Similarly, y = 2 was shown to be the equation of thfc
horizontal line whiph is parallel to and 2 units above the #-axis.
And this example illustrates the fact that an equation in y alone
represents a line parallel to the z-axis. Furthermore, it is proved
in analytic geometry that the graph of every first-degree, or linear,
equation in x and y is a straight line; and, conversely, that every
straight line is the crraDh of a linear equation.
170
Linear Equations and Graphs
Sec. 9-4
We shall proceed by first preparing a table of corresponding
values in a given problem and then plotting the corresponding
'points on the coordinate system to obtain the graph of the equation,
The following illustrative examples will point the way toward an
understanding of the procedure in the graphing of linear equations.
Example 9-4. Graph the function 2x + 3.
Solution: Let y = 2x + 3. Then assign any values for or, substitute them in the
equation, and obtain the corresponding values for ?/. The table and the graph are
shown in Fig. 9-1.
Since a straight line is definitely determined when two points are known, only two
pairs of values of x and y are needed in graphing a linear equation. We can, how-
ever, use three points in order to check our work.
y
3
5
-1
/U,5)
FIG. 9-1.
FIG. 9-2.
Example 9-5. Graph the equation 2x + 3?/ = 6.
Solution: The equation may be solved for y in terms of x. Then
y = - | + 2.
A table of corresponding values of # and ?/ and the graph are shown in Fig. 9-2.
9-5. INTERCEPTS
In general, the points where a curve crosses the coordinate axes
are the easiest to obtain.
The ^-intercepts are the values of x at the points where the graph
crosses the x-axis. Since y = on this axis, the ^-intercepts are the
values of x that correspond to y = 0. Similarly, the ^-intercepts are
the values of y at which the graph crosses the i/-axis. They are the
values of y that correspond to x = 0. Hence, we have the following
rule :
To find the ^-intercepts, set y = in the equation and solve for x.
To find the ^-intercepts, set x = in the equation and solve for y.
Sec. 9-6
Linear Equations and Graphs
171
Example 9-6. Find the intercepts of the line
2x + 3y = 6.
Solution: To find the ^-intercept, let y = 0. Then 2x = 6, and x = 3.
To find the ^/-intercept, let x = 0. Then 3y = 6, and y = 2.
Note that the intercepts a; = 3 and ?/ = 2 found in this solution correspond to
the points (3, 0) and (0, 2), respectively, where the line in Fig. 9-2 crosses the
coordinate axes.
9-6. GRAPHICAL SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS
In the graphical solution of two linear equations in two unknowns,
the graphs of the two equations are drawn with reference to the same
coordinate axes. Since the solution of two equations in x and y
is a pair of values of x and y which satisf^ both equations, the
solution must represent graphically a point common to both lines
represented by the equations. Hence, the values of x and y which
satisfy both equations give the coordinates of the point of inter-
section of the lines. We find, therefore, that the two lines intersect
in a single point, are parallel, or are identical, according as the
equations are consistent and independent, inconsistent, or consistent
and dependent.
Example 9-7. Solve graphically
2x + 3y = 24,
5x - 2y = 22.
Solution: Tables of corresponding values for the two equations and also their
graphs are shown in Fig. 9-3.
It is seen from the graphs that the lines intersect at the point (6, 4). That x = 6,
y = 4 gives the solution of the given equations may be checked by substitution.
Y
2x + 3y = 24 5x - 2y = 22
12
22
5
y
-11
FIG. 9-3.
172
I/near Equations and Graphs
EXERCISE 9-2
Sec. 9-6
In each of problems from 1 to 9, graph the given function. In each case give the
x- and ^-intercepts.
1. x - 3y = 1. 2. y = 2x - 8. 3. y - a? - 4.
4. = x -f 5. 5. = 3.r. 6. = #
7. y = 3z + 4. 8. y = x - 3. 9. ?/ = 3x - 5.
Solve each of the following systems of equations graphically.
10. fa? - 3// = 1,
\3x - 2y = 0.
13. (2*/ = a; + 3,
I2x = 16 - 3y.
16. \2x - y =4,
+2y = 12.
11. 130 - 2x = 0,
I x + y =2.
14. /3x - = 4,
i - 3x = 1.
12. (20 = * -3,
\2* = y + 3.
15.
-0
~r =4.
17.
+ 4 = 50 - 3,
- 4 =4x +2.
18.
- y)
2* - 30 = 12.
10
Deferminanfs
10-1. DETERMINANTS OF THE SECOND ORDER
Let us consider the following system of two linear equations in
two unknowns: , .
( aix + b\y - ci,
(10-1)
[ ClzX + b>2 y . C2.
The solution of these equations by the method of Section 9-1 is
given by
nn o\ k-*! "" t )lC2
(10 -2) x , --- p- > y
- - 2
0,201
It is understood that a^b 2 a< 2 bi ~ 0.
We shall at this point introduce a more convenient way of writ-
ing the expression (tib 2 Onbi. The notation which we select for
this purpose will enable us to express also the numerators of (10-2)
by means of the same symbol with the proper changes in letters.
In choosing a symbol we shall select a form which will exhibit
the numbers a l9 &i, a 2 , &n in the same relative positions as in (10-1) .
Thus, we write
02 62-
This arrangement of the four numbers in a square array, consist-
ing of two rows and two columns, is then enclosed within vertical
bars, as follows :
a i hi
a 2
This symbol represents a determinant of second order.
Thus, we start with a square array, or matrix, such as
174
Determinants
Sec. 10-1
We then associate with this array a number ai& 2 & 2 &i, called its
determinant, which is denoted in the following manner :
The numbers a u b lt a 2 , b 2 are called elements of the determinant.
The numbers ai and & 2 lie on the principal diagonal ; the numbers
a 2 and &i lie on the secondary diagonal.
Note. We observe that the "expanded" value of the foregoing
determinant is equal to the product of the elements on the principal
diagonal minus the product of the elements on the secondary
diagonal. It is interesting to note also that this value is the alge-
braic sum of all possible products obtainable by taking one and
only one element from each row and one element from each column.
Each product is preceded by a plus sign or a minus sign, according
to a rule to be stated in Section 10-2.
Using the notation of determinants, we can write the solution
(10-2) of (10-1) in the form
(10-3)
X =
y =
0,2
We note that the value of each of the unknowns in (10-3) may be
written as a fraction whose denominator is the determinant of the
coefficients as they stand in (10-1), and whose numerator is the
determinant formed from that of the denominator by replacing
the column of coefficients of the unknown in question by the column
of constant terms.
Note. If a 2 = kai and & 2 fc&i, where k is any number, then
= 0.
In this case, the equations of the system (10-1) are inconsistent
unless both numerators of the fractions in (10-2) are also equal to
zero, that is, unless
= and
= 0.
Therefore, the equations of the system (10-1) represent distinct,
parallel straight lines if c 2 = kc l9 or they represent coincident lines
if Co = kc\.
Sec. 10-2
Determinants
175
Example 10-1. Solve the system of equations
-.x + 3y = 1,
\x - 2y = 22.
Solution: Using determinants, we have
1 3
Also,
22 -2
4 3
3 -2
4 1
3 22
4 3
3 -2
- 2 - 66 - 68
- 8 - 9 - 17
= 4.
88-3
85
_ 8 - 9 - 17
= -5.
10-2. DETERMINANTS OF THE THIRD ORDER
A determinant of the third order is a number designated by a
square array of nine elements arranged in three rows and three
columns and enclosed within vertical bars. An example is
(10-4)
D =
a2 62
The value, or expansion, of the determinant (10-4) is defined
as the quantity
(10-5)
C3
63 C3
or as the quantity
(10 G) D = 0162^3 ai&3C2 ^
Here the products such as ai& 2 c 3 , ai& 3 c 2 , and
the terms of the determinant.
&2
63
are known as
Minors and Cofactors. In any determinant the minor of a given
element is the determinant of the array which remains after delet-
ing all the elements that lie in the same row and in the same
column as the given element. Thus, in (10-4) the minors of a x , 61,
Ci are, respectively,
C2
tt2 C2
63 3
6 2
&3
176
Determinants
Sec. 10-2
The cof actor of an element which lies in the z'th row and fcth
column is equal to the minor of that element if i + k is even, and is
equal to the negative of the minor if i + k is odd. That is,
cof actor = ( 1)*+* minor.
Thus, in the determinant in (10-4), the cof actor of &i equals the
minor of ai, since 0,1 lies in the first row and in the first column
and i + fc = l + l = 2, which is even. Similarly, the cofactor of bi
is the negative of the minor of 61, since i4-fc = l + 2 = 3, which is
odd.
Often the following procedure may prove more convenient for
finding the sign corresponding to a given element. Beginning with
+ in the upper left-hand corner, change sign from place to place,
moving horizontally or vertically, until the position for the element
in question is reached. The schematic arrangement of signs cor-
responding to the elements of a third-order determinant is thus
as follows :
Note that the sign for any position is independent of the path
followed in arriving at that position.
We shall designate the value of the cofactor of an element by the
corresponding capital letter, and we shall use the subscript that
occurs with the element itself. Thus, the cofactors of a t , 61, c x are,
respectively,
A, =
C2
(10-7)
a.3 C'3
Hence, (10-5) for the expansion of the determinant may also be
written as follows :
(10-8) D = aiAi + biBi + aCi.
This sum is called the expansion of the determinant according to
the elements of the first row.
We observe at this point that the right member of (10-6) repre-
sents all possible products, here 3 ! in number, that can be formed
from the determinant in (10-4) by taking one and only one element
from each row and each column. It follows also that the value
of the determinant is the same, regardless of the row or column
Sec. 10-3
Deferm/ncmfs
177
according to which the expansion is made. Thus, we may express
the determinant as
(10-9)
or as
D =
+ b 2 B 2 +
(10-10) D = 03^.3 + 63^3 + c^Cz.
These equations represent the expansion according to the elements
of the second column and according to the elements of the third
row, respectively.
Example 10-2. Expand the determinant
2-53
6
- 1
2 1
7 4
according to the elements of the first row and according to the elements of the
third column.
Solution: The expansion according to the elements of the first row is
2 1
7 4
6 1
- 1 4
6 2
- 1 7
This reduces to
2(8 - 7) + 5(24 + 1) + 3(42 + 2) = 259.
Expanding according to the elements of the third column, we have
() 2
-1 7
-I-
2 -5
-1 7
2 -5
= 3(42 + 2) - (14 - 5) + 4(4 + 30) = 259.
. PROPERTIES OF DETERMINANTS
From the definition of the value of a determinant we may deduce
the following important properties of determinants. These proper-
ties supply us with more convenient methods for evaluating a
determinant.
Note. For a more complete discussion of these properties, the stu-
dent is referred to any one of the various treatises on determinants
or to texts on the theory of equations or on solid analytic geometry,
where he will also find proofs which apply to determinants of
any order.
The properties listed here will be employed in examples that fol-
low, and their usefulness in simplifying determinants will be
illustrated.
178
Determinants
Sec. 10-3
Property 1. The value of a determinant is not changed if its rows
and columns are interchanged.
Property 2. If all the elements of a row, or of a column, are multi-
plied by the same number, the value of the determinant is
multiplied by that number. For example,
ka\ b\
ka,2 62
= k
0,2 62
Property 3. If two rows, or two columns, of a determinant are
identical or proportional, the value of the determinant is zero.
For example, let the first two columns be identical, as in the
determinant
CL\ Cii C\
D
Q<2 C2
Cl'3
Then, expanding according to the elements of the third column,
we have
0,2 02
D =
(73
= 0.
Property 4. The value of a determinant is not changed if we add
to the elements of any column (row) any arbitrary multiple of the
elements of any other given column (row) .
For example,
61
(12
C2
3
+ rib i 61 Ci
+ ribz 62 C2
The proof follows. Expanding according to the elements of the
first column, we find that
L + nbi bi c
0,2 + nb2 b2 2
0,3 + 7163 &3 ^3
ai 61 c\
0,2 b2 02
dz 63 C.3
The last determinant vanishes, since two columns are identical.
Sec. 10-3
Defermmcrnfs
179
Example 10-3. Evaluate the determinant
43-1
D =
5 1
2 4
Solution: By adding 2 times the elements of the first row to the elements of the
second row, we obtain
4 3
- 1
4 3
- 1
5 1
2
=
13 7
2 4
3
2 4
3
If now we add 3 times the elements of the first row to the third row, the determinant
becomes
4 3-1
13 7
14 13
Expanding according to the elements of the third column, we have
13 7
14 13
= -71.
In this example, we first converted the given determinant to one in which all but
one of the elements of the third column are zero. For the final expansion, the given
determinant was thus reduced to a determinant of the second order, and its value
was easily found.
Example 10-4. Without expanding the determinant, show that x = 2 satisfies
the equation
3x 2 x 3 -x
3 1
6 -4
= 0.
Solution: Substituting 2 for x in the determinant, we have
12 -8 2
3 1 7
6 -4 1
This equals zero, since the first and third rows are proportional. Hence, the
equation is satisfied by x = - 2.
180
Deferm/ncrnfs
Sec. 10-3
EXERCISE 10-1
In each of the problems from 1 to 12, evaluate the given determinant.
1.
5.
9.
2 -5
2 7
7 2
9 7
2.
3,
4.
.
3 -5
3 -5
14 4
8 14
1 -
-3 2
2
2
3-5
3 8
6.
7.
2
1 4
8.
3 -2
- 1
6-10
6 -6
-2
1 -3
2 1
4
02 1
x y 1
2
4 -5
226
- 1 3 2
10.
xi yi 1
- 11.
-6
- 1 4
12.
1 -6 3
34-5
X 2 02 1
4
8 -9
5 7 15
* In each of the problems from 13 to 19, solve the given system of simultaneous
equations by means of determinants. Check all solutions by substitution in the
equations.
13.
x + 30 = 5,
2x - 4y = 7.
14. |2x - y= 1,
\3x +2y =4.
15. f 3x + 20 = 5,
- 30 = 5.
16. If D represents the determinant in Problem 10, show that D is the equation
of the straight line through the points On, iji) and (j 2 , 2/2).
17. Find the ^-intercept and the 0-intercopt of the line whose equation is
x y I
4 1
= 0.
2 -3 1
18. Solve graphically the following system of equations:
x y I
x y I
2 1
-o,
1 1
1/2 1 1
1/2 1
= 0.
19. Find the coordinates of the vertices of the triangle whose sides are the straight
lines x - y + 2 = 0, 2x + 90 + 15 = 0, and 7x + 40 - 30 = 0.
20. Find the vertices of the parallelogram formed by the following lines:
x y 1
2 1 1
-3 2 1
= 0,
x y I
6-11
1
= 0,
x y I
1 3 1
2 4 1
= 0,
X
y i
5
i i
1
-3 1
= 0.
Sec. 10-4
Deferm/nanfe
181
10-4. SOLUTION OF THREE SIMULTANEOUS LINEAR EQUATIONS IN THREE
UNKNOWNS
Let us consider the following system of linear equations :
ix + biy + ciz = di,
(10-11)
C2Z =
The determinant of the coefficients of the unknowns is
i 61
D =
0,2
C2
For the solution of such a system, we employ the following theorem,
which is known is Cramer's Rule.
Theorem. If the determinant D of the coefficients of the system
is not equal to zero, the system has just one solution. In this solu-
tion, the value of any unknown is equal to a fraction whose denom-
inator is D and whose numerator is obtained from D by replacing
the column of coefficients of the unknown in question by the column
of constants di, d 2 , and c? 3 .
Proof. Let the numerators of the fractions for x, y, z be denoted
by DI, D 2 , D 3 , respectively. We proceed to show that if equations
(10-11) are to be satisfied, then
(10-12) Dx = Di, Dy = Z> 2 , Dz = D 3 .
Specifically, the equation for x will have the form
(10-13)
(12
C2
#3 63
C2
63 C3
Similar equations may be written for y and z.
To find x 9 the first equation of the system in (10-11) is multi-
plted by the cof actor At, the second by A 2 , and the third by A 3 .
After adding and collecting terms, we obtain
(10-14) (aiAi + a 2 ^ 2 + a 3 A 3 )x + (biAi + b 2 A 2 + bzAz)y
+ (ciAi + c 2 ^2 + c^A^z = diAi + ^2^2 + ^3^3.
The coefficient of x in (10-14) is the expanded value of D accord-
ing to the elements of the first column. The coefficient of y is
biAi + b 2 A 2 +
182
Defermmonfs
Sec. 10-4
this is equal to
b\
62
&3 &3 <?3
which equals zero, since two columns are identical. Similarly, the
coefficient of z is zero. Hence, we have shown that the left side of
(10-14) is the expanded form of the left side of (10-13) and that
the two are therefore the same.
The right side of (10-14) is cM-i + d 2 A 2 + d 3 A 3 , which is the
expansion of the determinant
i 61 i
This determinant may be obtained from D by replacing the coeffi-
cients of x by the column of constants ; that is, it is the expansion
of the determinant that we have called DI. Hence, the equation
Dx = DI in (10-12) is established. By a similar procedure we can
show that the equations Dy = D 2 and Dz = D 3 are also valid.
If D T^ 0, the value of x is given by the equation
d\ bi
(10-15)
x =
Similar values may be found for y and z. The proof is complete.
The proof of (10-12) is valid whether D ^ or D = 0. If D ^ 0,
the equations of the system (10-11) are consistent and have only
one solution, which is of the form (10-15) ; that is,
(10-16)
= 5"' v = ^'
If D = 0, and any one or more of the other determinants, D\, D 2 ,
DZ, is not zero, the given system of equations has no solution and is
inconsistent.
Sec. 10-5
Defermmcrnfs
183
If D = 0, and all the other determinants are zero, the equations
of the system (10-11) may be consistent or inconsistent. If they
are consistent, there are infinitely many solutions. This case will be
treated in Section 10-5.
The following example will illustrate the case for which D ^ 0.
Example 10-5. Solve the system of equations
x - y + z = 1,
x + y - 2z = 3,
2x - y + 3z = 4.
Solution: Here
D =
1
-1
1
1
1
-2
2
- 1
3
1
1
1
1
3
-2
2
4
3
1
- 1
1
3
1
-2
4
- 1
3
1
- 1
1
1
1
3
2
- 1
4
Hence, x = -~ =
y ~ D ~5
= 11;
= 2.
= ^ = - If we check by substitution, we
j_J O
find that these values satisfy the given equations.
10-5. SYSTEMS OF THREE LINEAR EQUATIONS IN THREE UNKNOWNS WHEN
D =
We note that when D = 0, the system (10-11) will not have a
solution if any one of the other determinants DI, D 2 , D 3 is different
from zero. Suppose that a solution is given by
x = r, y = s, z = t.
Then the equations (10-12) become
r = Z?i, s = D 2 , * = D 3 .
It follows that Z?i = 0, Z> 2 = 0, Z> 3 = 0.
The following example will illustrate the case of consistent equa-
tions where D = and Z?i = D 2 = #3 = 0. The equations are said to
be dependent, and they have infinitely many solutions. The student
should construct an example to show that the equations (10-11)
may be inconsistent when D = 0, even though Z>i = D 2 = D 3 = 0.
184
Determinants
Sec. 10-5
Example 10-6. Solve the system of equations
x + y - z=3,
2x - y + 30 = 1,
4z - 2?/ + 62 = 2.
Solution: Here
Z) =
1
1 - 1
2 -1
4 -2
This equals zero because the second and third rows are proportional. But, we also
find that
1 1 3
D l =
3 1-1
1-1 3
2-2 6
13-1
2 1 3
42 6
2-11
4-22
= 0.
In this case the given equations have a solution. In fact, the second and third
equations are proportional, and so either of those can be solved together with the
first equation for two of the unknowns in terms of the third. For example,
4 - 2z
5 -f 5z
x =
Thus, we have a single value of x and a single value of y for every value of z.
However, there are infinitely many values of z, and therefore infinitely many
solutions of the given equations exist.
10-6. HOMOGENEOUS EQUATIONS
The system (10-11) is homogeneous if di = 0, d 2 = 0, and d 3 = 0.
Such a system always has the trivial solution x = y = z = 0. When
di = d 2 = d 3 = 0, it is seen that DI = D 2 = D& = 0, for each of these
determinants has zero for every element in one column. If D = 0,
it follows from (10-16) that we can have but one solution, which
is given by
Hence, if the given system is to have a solution besides the trivial
solution, D must equal zero. It may be shown that, if D = 0, non-
trivial solutions always exist.
Example 10-7. Solve the system of equations
x - y + 2=0,
2x 3y + 4z = 0,
K/p _ 2?7 - 9. " f)
See. 10-7
Determinants
185
Solution: The determinant D is
1
- 1
1
- 1
2
-3
4
=
- 1
-3
1
5
-2
- 1
3
-2
-3
= 0.
Therefore, nontrivial solutions exist. To find these solutions, we proceed as follows:
Transpose z in each of the first two equations, and solve for x and y in terms of 2.
Then we have
D =
1 - 1
2-3
z
- 1
-42 -3
= - 2,
1 -2
2 -42
Hence,
I j 2
= -p = 2, and i/ = =
Substitution shows that those values also satisfy the third equation. The given
system therefore has infinitely many solutions, and the values of x, y, and z are re-
lated by the equations x = z and y = 2z.
EXERCISE 10-2
In each of the problems from 1 to 6, solve the given system of simultaneous
equations by means of determinants. Check all solutions by substitution in the
equations.
1.
4.
3x + y - 2 = 11,
x +3y - z = 13,
3 + y -3z = 11.
2.
= - 10, 5.
2x - 20 + 2-2,
6x + 2y - 2z = 5.
x - y - 22 = -1, 3.
5x -2y = 0,
I2x - 4y + z=3.
2x - y - 32 = 7, 6.
x + 2y - 2 = 10,
3jc - 3# + 22 = - 7.
x - ?/ + 62 = 7,
2z + 3y + 62 = 0,
3 + 2y + 92 = 3.
2x - y = 3,
2.r - 32 = - 1,
32 - y ~ 2.
For each of the following systems find at least one nontrivial solution, or show
that there is no nontrivial solution.
7.
2x -
+ 22 = 0,
+2=0,
- 2=0.
8.
2x - 3y 4- 42 = 0,
x +3y - 2=0,
7z + 3y + 62 = 0.
9. [ x - 2y + 32 = 0,
J2.c - y +42 =0,
3x + 2/ - s = 0.
10-7. SUM AND PRODUCT OF DETERMINANTS
.Closely related to Property 4 of Section 10-3 is a theorem con-
cerning the sum of determinants. We shall illustrate the theorem
for splitting the elements of a given column into two parts by
means of the following equality for third-order determinants :
ai + 61 ci
0,3 + 62 02
03 + &3 Cs
d\
0,2 2
as 03
d\
62
186
Deferm/ncmfs
Sec. 10-7
This can be shown to be true by expanding the three determi-
nants according to the elements in the first column and noting that
in the expansion the minors are the same for all three determinants.
Example 10-8. Show that
2 45
1-10
3 76
+
1 4 5
4-10
- 1 76
=
3 45
5-10
2 76
Solution: Expanding each of the determinants on the left side of the equation
according to the elements in the first column, we have
- 1
4 5
4 5
- 1
4 5
4 5
+ 3
+
-4
7 6
7 6
- 1
7 6
7 6
- 1
- 1
4 5
4 5
___ o
-5
+ 2
7 6
7 6
- 1
This result is the expansion of the determinant on the right in the given equation
according to the elements of the first column. Computing the value of each determi-
nant in the given equation, we see that each side reduces to 47.
A similar theorem for splitting the elements of a given row is illustrated by the
following example :
ttl + Ci 61+6
(12
CL2 62
This can be shown to be valid by expanding according to the elements of the
first row.
Thus, consider the determinant
4 5
1 7
This may be written as a sum in various ways. Examples arc
4 5
and
2
2 3
1 7
2 2
1 7
-2
4 5
3 5
We shall state without proof the rule for the product of two determinants of
the same order. The product is equal to a determinant of like order in which the
element of the ith row and kth column is the sum of the products of the elements
of the ith row of the first determinant and the corresponding elements of the kth
column of the second determinant. For example, for second-order determinants,
we may write
1-5 +2-7 1-6 +2-8
1 2
3 4
5 6
7 8
3-5 +4*7 3*6 +4*8
Sec. 10-7
Deferm/ncrnfs
187
That the product of the values of the two determinants on the left equals the
value of the determinant on the right is checked by expanding. Thus, the desired
equality becomes
(4 - 6) (40 - 42) = (950 - 946),
or
(-2). (-2) = 4.
To illustrate the multiplication of two third-order determinants, we have that
3
2 5
1
7 6
1
- 1 2
- 3
9 3
4
6
5
-4 2
is equal to
3-1 +2(-3) +5-5
3-7+2-94- 5( - 4) 3-6+2-3+5-2
1-1 +(-!)( -3) +2-5 l-7 + (-l)9+2(-4) 1 6 + (- 1)3 +2 2
.4-1+ 6(- 3) +0-5 4. 7+6-9+ 0(- 4) 4-6+6-3 +0-2
This reduces to
22 19 34
14-10 7
- 14 82 42
which is equal to 630. Also, computing the values of the given factors, we have
30(- 21), which equals - 630.
EXERCISE 10-3
In each of the first three problems, combine the given determinants into a single
determinant, and evaluate the result.
1.
3.
022
2 2-7
-3-1 7
+
-3-1 7
4 8-10
4 8-10
122
1 4 2
2-3 1
-
22 1
4 8-1
4 19 - 1
2 4-5
2 4-5 2 4-5
-6-1 4
+
_ 6 _ 1 4 + _ 6 - 1 4
1 2 12
4 3-6 -1 3-16
188
Determinants
Sec. 10-7
In each of the following problems, find the product of the determinants without
evaluating the individual factors.
4.
8.
3
1
3 1
1
2
3-2
.
.
5.
.
1 -2
1 5
2
-
1
5 4
1 2
2 3
1
3
- 1 4
.
.
7.
.
.
2 3
3 1
5
1
3 10
(1
2
3 -1 \ 2
6
-1
4
5 1 / -3
8
c
-2
3
4 5 \ / 3 2
6 7 / \ 5 4
-3 2
1 4
)
11
Complex Numbers
11-1. THE COMPLEX NUMBER SYSTEM
There are many problems that cannot be solved by the use of
real numbers alone. We observe, for example, that the equation
x 2 + 1 = has no real root, since x 2 can never be negative if # is a
real number. In order to provide solutions to such equations, a new
system of numbers, called the complex number system, was intro-
duced. Later in this book, we shall find many instances of solutions
involving complex numbers.
We shall now define a complex number as an ordered pair of real
numbers, which we denote by (a, b). If the numbers a and b are
regarded as the Cartesian coordinates of a point in a rectangular
coordinate system, we have a one-to-one correspondence between
the set of complex numbers and the set of points in a plane. The
plane is called the complex plane. Two complex numbers (a, b)
and (e, d) are equal if and only if they correspond to the same
point, that is, if and only if a - c and b - d.
Addition, subtraction, and multiplication of complex numbers
are defined as follows :
(11-1) (a,6) + (c,d) =(a + c,6 + d)i
(11-2) (a, 6) - (c, d) = (o - c, 6 - d),
(11-3) (a, 6) (c, rf) = (ac - 6d, ad + be).
For example,
(1, 3) + (5, 2) = (6, 5),
(1, 3) - (5, 2) = l- 4, 1),
(1,3). (5, 2) = (-1,17).
We also define the following special complex numbers :
(11-4) 0- (0,0),
(11-5) 1 - (1, 0),
(11-6) i = (0, 1).
189
1 90 Complex Numbers Sec. 1 1-1
The complex number serves as a zero of the complex number
system, while 1 serves as a unit, in Accordance with the following
properties :
+ (a, 6) = (a, 6) + = (a, 6),
- (a, 6) = (a, 6) = 0,
1 (a, 6) = (a, 6) 1 = (a, 6).
The so-called imaginary unit i= (0, 1) will be discussed in more
detail in Section 11-2.
If A; is a real number, we define
(11-7) * (a, 6) = (fc, 0) (a, 6) = (fca, kb).
Also, we define
(11-8) - (a, 6) =(-!) (a, 6) = (- a, - 6).
Since (a, 6) + (a, 6) = (0, 0), the complex number (a, b) is
called the negative of (a, 6).
The Reciprocal of (a,b). If (a, &) ^ 0, then it has a reciprocal
(x, y) such that
(a, 6) (x, y) = 1.
Furthermore, the reciprocal is given by
(11-9) (x , y)
^
By (11-3) and (11-5), the equation (& &) '(#,?/) =1 may be
written in the form
(ax - by, ay + bx) = (1, 0).
Since a 2 + & 2 ^ 0, we may determine x and T/ by solving the simul-
taneous equations
j ax - by = 1,
1 6x + ap = 0.
The values of x and y can be foundry any of the methods taken up
in Section 9-1. The solution is
a _ b
X ~a 2 + b*' y ~ a 2 + 6 a <*
We have, therefore, verified (11-9) .
Division of (a, b) by (c y d). If (c, d) = 0, division can be defined
as follows :
(a, 6) -5- (c, d) = (a, 6) (u, v),
where (u,v) is the reciprocal of (c,d). By (11-9), we have
(11-10) (a, 6) -!- (c, d) = (a, 6)
(
ac + bd ad + bc
Sec. 1 1 -2 Complex Numbers 1 9 1
For example,
(5, 13) + (3, -2) = -,
This result can also be verified by the method in Section 11-3.
11-2. THE STANDARD NOTATION FOR COMPLEX NUMBERS
The special number i - (0, 1), defined by (11-6), has the follow-
ing property :
(11-11) i* = (0, 1) (0, 1) = (- 1, 0) = - (1, 0) = - 1.
We shall now show that (a, 6) and the binomial form a 4- bi are
equivalent, or that
(11-12) (a, 6) = a + bi,
in which a + bi means al + bi. By (11-5), (11-6), and (11-7),
al + bi = a(l, 0) + 6(0, 1) = (a, 0) + (0, 6).
Finally, by (11-1), we have
(a, 0)4- (0,6) = (a, 6).
Hence, (11-12) is established.
Real and Imaginary Parts of Complex Numbers. We call a the
real part and & the imaginary part of the complex number a + bi.
If a and 6^0, a + bi reduces to bi, which is called a pure
imaginary number. If 6 = 0, the complex number a + bi, or al 4- bi f
reduces to the complex number al, which may be identified with
the real number a. The complex numbers, then, include both the
real numbers and the pure imaginary numbers as special cases.
Illustrations of various classes of numbers follow :
Some real numbers are 2, 5, and \/3.
Some pure imaginary numbers are 3i, and x/Bt .
Some complex numbers are 2 + 3i, and \/S i.
Note that the numbers 2, 5, \/3, 3f, and \/5i may be put into
the standard form a 4- bi and written, respectively, as the complex
numbers -2 + Of, 5 4- Of, \/3 4- Oi, + 3f, and - \/5f. Since
= + Of, which may be written briefly as 0, we shall drop the use
of bold-face 0; similarly for bold-face 1.
Conjugate Complex Numbers. The conjugate of a complex num-
ber a + bi is defined as a bi. Likewise, a + bi is the conjugate of
a bi. Some pairs of conjugate complex numbers follow: 2f, 2f ;
3 + 5f , 3 5f ; and x + 2yi, x 2yi . A real number is its own
conjugate.
192 Complex Numbers Sec. 11-2
Powers of t. It is readily seen that i 3 = i 2 i = i, i* = i 2 i 2 = 1,
i = ft i = i 9 6 ft . 2 = _^ an( j so on Therefore, successive posi-
tive integral powers of i have only four different values, namely,
i, 1, i, and 1 ; these four values are repeated in regular order.
Hence, if n is any positive integer, we have in general
These relationships afford a simple method for evaluating powers
of i, as shown by the following illustrations : i 7 = i* +3 i 3 = i ;
i 38 = &*+* = i 2 = -1 ; and i 103 = i 4 ' 26 - 1 = i.
11-3. OPERATIONS ON COMPLEX NUMBERS IN STANDARD FORM
From the definitions given in Sections 11-1 and 11-2, it follows
that a 4- bi and c + di can be added, subtracted, multiplied, and
divided as if they were real binomials, except that, where i 2 appears,
it is replaced by 1.
Algebraic Addition and Subtraction. Addition of two complex
numbers is effected by adding their real and imaginary parts sepa-
rately ; and subtraction is performed by subtracting their real and
imaginary parts separately. Thus, in accordance with (11-1) and
(11-la) (a + bi) + (c + di) = (a + c) + (b + d) i,
and
(ll-2a) (a + bt) - (c + di) = (a - c) + (b - d) i.
For example,
(3 + 20 + (4 - 50 = (3 + 4) + (2 - 5)i = 7 - 3t,
and
(3 + 20 - (4 - 50 = (3 - 4) + (2 + 5)i = - 1 + 7i.
We note that the sum of conjugate complex numbers is a real num-
ber, because (a + bi) + (a bi) = 2a. Also, the difference of two
conjugate complex numbers is a pure imaginary number, because
(a + bi) - (a - bi) = 2bi.
Algebraic Multiplication. To find the product of two complex
numbers, multiply them according to the rules of algebra, and
replace i 2 by -1 in the result. Thus
(a + 60 ( c + di) = ac + adi + bci + bdi 2 .
Replacing i 2 by -1, we have, in agreement with (11-3),
(ll-3a) (a + bi) (c + di) = (ac - bd) + (ad + bc)i.
Sec. 11-3 Complex Numbers 193
In many respects the notation a + bi is more convenient than
(a, 6) . In particular, the former notation makes it easier to remem-
ber how to multiply two complex numbers. For example,
(3 + 20 (4 - 5z) = (12 + 10) + (- 15 + 8)i = 22 - 7t.
The student should note that the product of two conjugate complex
numbers is a non-negative real number, because (a 4- bi) (a bi) =
a 2 + b 2 .
Algebraic Division. To obtain the quotient of two complex num-
bers, multiply the numerator and the denominator by the conjugate
of the denominator. Thus, if c + di ^ 0,
a + bi __ a + bi c di __ ac adi + bci bdi 2
c + di ~~ c + di c di ~~ c 2 d 2 i 2
_ (ac + bd) + (be - ad)i
c 2 + d 2
Therefore, . ,. , , , , ,
a + bi __ ac + bd be ad .
'c~+~di ~ c 2 + d 2 ~^ c 2 + d 2 lm
The right member is of the form A + Bi, and this equation agrees
with (11-10).
;
Example 11-1. Reduce - to the form a + bi.
i
Solution: The conjugate of i is i. Then
Example 11-2. Find the value of 5 + 13i divided by 3 - 2i.
Solution: Represent the division as ^ -~ > and multiply the numerator and
the denominator by 3 + 2i. We get
5 + 13i 3 +2i __ (15 - 26) + (39 + 10)i _ 11 49 .
3 - 2i ' 3 + 2i ~ 9+4 "" 13 + 13 *'
EXERCISE 11-1
In each of the problems from 1 to 12, express the given quantity in the form
a + bi and give its conjugate. In working these problems, note that V a *
l. v^ 7 ^. 2. V 17 ^. 3. - V
4. V- ^' 2 . 5. - V- 3(xi 2 . 6. V- a?
7. 3 V2 + 3 V- 2. 8. 4 - 3 V- 16. 9. 1 + 2 V- 8.
10. \A 2 ~ V- * 2 - 11. V15 4- V- 64a 3 6. 12. 3 + V- 32a 2 6 3 .
194 Complex Numbers Sec. 11-3
In each of the problems from 13 to 26, compute the value of the given expression.
13. i. 14. i 12 . 15. (-i) 13 . 16. - (-i) 17 .
17. i". 18. -i 70 19. (-i) 235 . 20. (-i) 602 .
21. i s i 29 . 22. i 37 i 183 . 23. i 14 - (- i) 18 . 24. i 10 + i ao + i 30 .
25. i 25 - i 50 + i 75 - i 100 . 26. i 10 -f i 100 -f i 1000 -f i 10000 .
In each of the problems from 27 to 36, find the values of x and y which satisfy the
given equation.
27. (2z, 3y) = (3, 1). 28. (2*, 3?) = (8, 9).
29. (to, 5y) = (18, - 25). 30. (x + y), to + 1) = (x, 1).
31. (to +2y,x + 5y) = (20 + 3, - 19).
32. z + (y - x)f = 1 + 3f. 33. 3z - 6 - (5 - 20)i = 0.
34. 2z - 40i = 6 - 2st. 35. 3x - 7 = (4 - 30)i.
36. 3-ct - 2yi + 8x + 50 - 12 = 2x + 2yi + 5y + 6 + (x - 2)i.
In each of the problems from 37 to 78, perform the indicated operations and
reduce to the form a + bi.
37. (2, - 3) + (5, 6). 38. (1, - 1) + (3, 2). 39. (1, V) + ' ~
40. (1, - 7) - (7, - 3). 41. (2, 3) - (1, 1). 42. (1, m
43. (4, - 3) (4, 3). 44. (0, - I) 3 . 45. (1, 0) (0, 1) - (1, 1).
46.(0,1) 4 . _ 47
5 ' - ^~) 3 ' 49. V^ + V^2 - V9.
50. V" 17 ^ - V^2 - a . 51. (4 + 3i) + (2 - 3t).
52. ( - 2 + 3i) - ( - 6 - 3f). 53. (8 + 9i) + (5 + 2i).
54. (3 - 2i) - (3 + 5i). 55. ( - 3 + 2i) - (5 - 2i).
56. (6 + 2i) -f (3i + V" 17 !). 57. (2i + 3) + (8 - 5 V 17 !).
58. (6 + 3i) 4- (3 - 5i). 59. (2 + 5i) (6 - 3i). 60. 2f (5 + 3i).
61. (5 -f- 1) (6 -f 2i). 62. (3 - 3i) (4< +2). 63. (1 + i V2) (5 -f- 2f).
64. (5< - 8) (2i - 4). 65. (5 - 3t) (2 - 4i). 66. (3 - f) -5- (2 - 5i).
67. (3< + 4) + (1 - i). 68. (2 + i) 2 + (5 - f). 69. (2 - 3t) -s- (5 - 4<).
70. (3 - V3i) -s- (4 + 5i) 2 .
71. (3 - v/3i) 2 + (3 + V30 2 - (4 - 3i) (i - 6).
72. (6 - 2i) (1 -f f) (1 - 3i) + (4 - 5i).
73. 1 + (6 - 5t). 74. (6 - 5i) *- 1. 75. i 4- (4 - 3i).
76 3 -4i j^+j)^^^ (3 -I- 5i) (8 + Gi)
IU ' (2 + i) (3-2i) "' (6 - W) (2i - 7) '' (4 - 7) (4 + ) '
79. Prove that complex numbers satisfy the associative and commutative laws of
addition and multiplication and the distributive law.
80. Prove that if (a + bi} (c + di) = 0, then a + bi = or c + di = 0.
Sec. 11-4
Complex Numbers
195
11-4. GRAPHICAL REPRESENTATION
As we have seen, the complex number a + bi determines a definite
point P in the plane whose rectangular coordinates are x = a and
y ~ 6. Conversely, to every point P in the plane corresponds a com-
plex number a + bi for which the values of a and b are the respec-
tive rectangular coordinates of P. See Fig. 11-1. In this sys-
tem, the real numbers a + Oi are represented by points on the
#-axis, which is called the axis of reals. Pure imaginary numbers
+ bi are represented by points on the 2/-axis, which is called the
axis of imaginaries.
a
FIG. 11-1.
FIG. 11-2.
It is more convenient at times to represent the complex number
a + bi by the vector drawn from the origin to the point P. The
length of the vector is given by the relationship r = V a ~ + 6 2 , and
the direction is given by an angle determined from the equations
a - r cos 6 and b r sin 6.
In Fig. 11-2 is indicated the graphical addition of the two com-
plex numbers a + bi and c + di, which may^ be represented either
by the points P l and P 2 or by the vectors OJ?i and OP^ Withjths
completion of the parallelogram OPiP-JP^ the sum of OP l and OP 2
can be represented by the diagonal OP 3 . Thus, either P 3 (a-f c,
6 + d) or OP l3 represents the sum (a 4- c) + (6 + d)L Hence, the
vector which represents the sum of two complex numbers is the
sum of the vectors representing the given numbers.
To subtract c + di from a + bi graphically, we merely add a + bi
and c di.
Example 11-3. Add the complex numbers 2 + 3f and 6 + 2t graphically.
Solution: Let Pi (2, 3) represent the number 2 + 3f and let P 2 (6, 2) represent
the number 6 + 2i. Draw OPi and OP 2 in Fig. 11-3, and complete the parallelo-
gram OPiPaPa. Then Ps^represcnts the sum 8 + 5i of the complex numbers
2 + 3i and 6 + 2i, and OP 3 represents the sum of the vectors OPi and OP 2 .
196
Complex Numbers
Sec. 11-4
/' 3 (8,5)
o
FIG. 11-3.
The difference of two complex numbers may be obtained in the
same manner if we apply the relationship
(a + to) - (c + di) = (a + bi) + (- c - di).
Let P and Q represent the numbers a + bi and c + di, respectively,
in the complex plane, as shown in Fig. 11-4. Then c di is repre-
sented by Q', which is the reflection of Q through the origin.
Q fad)
O
P(a,b)
FIG. 11-4.
Let us recall how the difference of two vectors was explained in
Section 6-7 and was represented graphically in Fig. 6-15. If we let
the vectors OP, OQ,jand OQ' represent a 4- bi, c + di, and -c -di,
respectively, then OR is the desired vector and R is the point
representing the number (a + bi) (c + di) .
11-5. TRIGONOMETRIC REPRESENTATION
Let the complex number a + bi be represented by the radius
vector drawn from the origin to the point P. Then the distance
\OP\ = r is called the modulus, or^the absolute value, of the complex
number ; and the angle 0, which OP makes with the positive #-axis,
is called an argument, amplitude, or angle of the complex number.
Sec. 11-5
Complex Numbers
197
From Fig. 11-1, it is clear that - = cos 6 and - = sin 0, or
a = r cos 8 and b = r sin 6.
Hence,
a + bi = r cos + (r sin 0)i = r(cos + i sin 0).
This last expression is known as the polar form or the trigono-
metric form of the given complex number, as contrasted with the
standard or rectangular form a + bi.
To reduce a given complex number a + bi to the trigonometric
form r(cos 9 + i sin 0), we find r and by means of the relation-
ships r = V^ 2 + b 2 , a - r cos 0. and 6 = r sin 0. We have
a + bi = rf - + - i) = r (cos + i sin 0).
Example 11-4. Represent the complex number
/5
~
6<r
V3/2
1/2
FIG. 11-5.
9 + ^r i graphically, and change the given notation
to the trigonometric form.
Solution: The point P whose rectangular coordinates
/I \/3\ , ., , 1 , \/3 . .
are [ - > ^- ) represents the number - + zr * Since
\J ^ / Z i
a = 1/2 and 6 = \/3/2 are both positive, ^ is a first-
quadrant angle. We see from Fig. 1 1-5 that r =
j^l- The angle 6 is
determined from the equations cos 6 = 1/2 and sin 6 = \X/2. In this case we may
let 9 = 7T/3 or 60. Hence, 1 + ^1 f = l (i + ^ f) = C os 60 + i sin 60.
J ^ \J ^ /
Example 11-5. Express the complex number 1 i in the trigonometric form.
Solution: Here a = 1 and 6 = 1. So r = \/2, and is a fourth-quadrant angle
determined from cos 6 = ^ and sin = -- y= We thus have
V2 -\/2
-4= - --Ut = \/2(cos 315 + i sin 315).
EXERCISE 11-2
In each of problems from 1 to 12, represent the complex number and its con-
jugate graphically.
1. 3 + 2f. 2. 8 + 2i. 3. 3 + . 4. 2 - 3t.
5. 3 - 5i. 6. 1 - 1. 7. i. 8. 1.
10.
11. (1 -
12. 5 + 12.
1 98 Complex Numbers Sec. 1 1-5
In each of the problems from 13 to 24, perform the indicated operations graph-
ically. Then check the result algebraically.
13. (7 - 30 + (- 4 + 0- 14. (2 + 30 - (4 - 5i).
15. (3 - 60 - (4 + 30. 16. (6 - 2i) + (6 + 20-
17. (3 + 20 - (5 - 0. 18. (3 + 40 - ( - 2 - 40.
19. (5 + + (1 - 50. 20. (3 - 2i) - (5 - 40.
21. 3 - (1 - 40 - (2 + 0- 22. (v/2 + + (1 + V&) - 7i.
23. 7 - (4 - 20 - (- 2 + i V). 24. (4 + 30 - (2 + 30 - (3 + 20-
In each of problems from 25 to 36, change the complex number to the trigono-
metric form and represent it graphically.
25. 1 + i. 26. - 5. 27. - 3t. 28. 3 - 3 \/3i.
29. |(1 + V20. 30. ~(1 - V30- 31. 5 + 12t. 32. -?
QQ 6 -4i 1 -i , ,, } ,., 2 Q(5 1 Q , 3 -2i
* -2TT -3-+!' d4 ' W ~ 4l) ' ** 6~+' 2 +< *
11-6. MULTIPLICATION AND DIVISION IN TRIGONOMETRIC FORM
Let TI (cos a + i sin a) and r 2 (cos /3 + i sin /?) be any two com-
plex numbers in trigonometric form. Then, their product is given
by the relationship
TI (cos a + i sin a) r% (cos /3 + i sin /3)
= fi^Kcos a cos /3 sin a sin /3) + i(sin a cos /3 + cos a sin ]8)]
= rir2[cos (a+ ]8)+ isin (a+ j8)J.
Thus, we have proved that the absolute value of the product of two
complex numbers is the product of their absolute values, and an
angle of the product is the sum of their angles.
The result found for the product of two complex numbers can be
extended to the product of three or more complex numbers.
The quotient obtained by dividing the complex number
1*1 (cos a + i sin a) by the complex number r 2 (cos/J + isin/3) is
given by the relationship
ri(cosa + fr'sinoQ _ ri(cos a + i sin a) cos |8 i sin ft
r2(cos ft + i sin ft) ~~ r2(cos ft + i sin ft) cos ft i sin ft
= [cos (a - ft) + i sin (a - ft)].
It follows that the absolute value of the quotient of two complex
numbers is the quotient of the absolute values, and an angle of the
quotient is the difference of their angles.
Sec. 11-7 Complex Numbers 199
Example 11-6. Find the product of 2(cos 30 + i sin 30) and 3(cos 120 +
f sin 120).
Solution: By the rule for products in polar form, we have
2(cos 30 + i sin 30) 3(cos 120 + i sin 120)
= 2-3 [COP (30 + 120) + i sin (30 + 120)]
= 6 [cos 150 + i sin 150] = 6 ( - ^~ + ) - 3 ( - V3 + i
i /^i
Example 11-7. Find the quotient when 1 i is divided by + -^- f.
Zi A
Solution: From Examples 11-4 and 11-5 in Section 11-5,
1 - i = \/2(cos 315 + i sin 315),
and
i + ~ i = cos 60 + i sin 60.
1 - i
i _ ft (cos 315 + i sin 315)
1 \/3 . cos 60 -f i sin 60
2 + 2 l
= V2 [cos (315 - 00) + i sin (315 - 60)]
= V2 [cos 255 + i sin 255]
= - \/2 (cos 75 + i sin 75).
From a table of trigonometric functions, we find that the result is
- V2 [0.2588 + i (0.9659)].
Or, using exact values of cos 75 and sin 75 previously found, we obtain
r - _ -i
- V2 \~ (V3 - 1) + i ^ (VS + 1) = - | [(V3 - 1) + i (V3 + 1)].
L. Hb 4: _J Z
11-7. DeMOIVRE'S THEOREM
If we extend the law of multiplication of the preceding section to
n factors, we have
[ri(cos di + i sin 0i)] [r2(cos 02 + i sin #2)] [r w (cos n + i sin n )]
= rir 2 - - - r n [cos(0i + 2 + - - + n ) + i sin (0i + 2 + - - - + 0*)].
If how we put TI = r 2 = = r n = r and = 2 = ' = O n = 0, it fol-
lows that
[r(cos 6 + i sin 0)] n = r n (cos n0 + i sin n0).
This result is known as De Moivre's Theorem.
Although we have derived De Moivre's Theorem only for integral
values of n, it can be shown to hold for all real values of n, if prop-
erly interpreted.
200 Complex Numbers Sec. 11-7
Example 11-8. Find the value of (1 i) 4 by De Moivre's Theorem.
Solution: Since 1 i = A/2 ( 7=- -p-i ] > the polar form of 1 i is
\\/2 V2 /
\/2 (cos 315 + i sin 315). Hence, by De Moivre's Theorem,
(1 - i) 4 = [\/2 (cos 315 + i sin 315 )] 4
= (V^) 4 [cos (4 - 315) + i sin (4 315)]
= 4(cos 1260 + i sin 1260)
= 4(cos 180 + i sin 180) = - 4.
Example 11-9. Derive formulas for cos 20 and sin 26 by De Moivre's Theorem.
Solution: By De Moivre's Theorem for n = 2, we have
cos 20 + i sin 20 = (cos +i sin 0) 2 = cos 2 + (2 cos sin 0) t - sin 2 0.
The two sides are equal only if the corresponding real and imaginary parts are
equal. Hence,
cos 20 = cos 2 - sin 2 0,
and
sin 20 = 2 sin cos 0.
11-8. ROOTS OF COMPLEX NUMBERS
Let p(cos < + tsin$) be an nth root of the complex number
r (cos + i sin 0), where ^ ^ 360. Then
[p(cos $ + i sin <)] n = r(cos 6 + i sin 0).
By De Moivre's theorem, this leads to
(11-13) p n (cos n<f) + i sin rup) = KCOS + i sin 0).
Our problem now is to find all non-negative numbers /> and all
angles <f> for which (11-13) is satisfied. Separating real and
imaginary parts, we have
(11-14) p n cos ncf) = r cos 0, p n sin n< = r sin 0.
Squaring and adding, we obtain
p 2n (cos 2 n<t> + sin 2 w$) = r 2 (cos 2 6 + sin 2 6).
Therefore, p 2 * = r 2 , since cos 2 a + sin 2 a = 1. The absolute value p
is then given by the equation
(11-15) p = ^
From (11-14) we then have
cos n<j> = cos 0, sin n< sin 0.
It is clear from these equations that the angles n$ and 6 can differ
only by a multiple of 2ir or 360. More precisely,
ft If .
(11-16) n<f> = + k 360, or = - +
7i
where 'fc is any integer.
Sec. 11-8
Complex Numbers
201
For fc = 0, 1, 2, , (n-1) in (11-16), we obtain n distinct
values of the angle, all of which are non-negative and less than 2?r
or 360. Corresponding to these angles we obtain n distinct roots
given by the formula
(11-17)
cos
/b-360
n
. . .
J- 1 sin
fc 360
n
For example, for k 0, we obtain one nth root, called the prin-
a
cipal root, with absolute value r 1/n and angle - ; for k = 1, we have
n
a second root, with absolute value r 1/n and angle
so on to k = n 1. The value k = n
would yield the same root as k 0,
since
n.860* 6
n
=
n '
/ fl \
cos (- + 360) = cos - and
\n / n
.n ^ D
sin (- + 360) = sin - Similarly,
\Tb f it
k n + 1 yields the same root as
k 1, and so on. This means that
only n distinct roots exist.
It is interesting to note that the
points which represent the roots
are equally spaced on a circle
whose radius is <\/r and whose cen-
ter is the origin. This fact is illus-
trated in the following example
and Fig. 11-6.
+ 360
n
; and
FIG. 11-6.
Example 11-10. Find the three cube roots of 8(cos 60 + i sin 60).
Solution: The three cube roots are found by evaluating (11-17). Thus, we have
360>
60 + i sin 60) =
60 + k 360 , . . 60 + i
- + i sin -
3 3
= 2[cos(20 + k 120) + i sin (20 + k 120)].
As just shown, the substitution of = 0, 1, 2 yields the three required roots.
Hence, for k = 0, we have 2(cos 20 + i sin 20) ; for k = 1, the root is 2(cos 140 +
t sin 140) ; and for k = 2, the root is 2(cos_260 + i sin 260). The roots are repre-
sented by the equally spaced vectors OPi, OP 2, and OPs, terminating on the circle
whose radius is 2 and making angles of 20, 140 J , and 260, respectively, with the
positive z-axis.
202 Complex Numbers Sec. 11-8
EXERCISE 11-3
In each of the problems from 1 to 18, perform the indicated operations by first
expressing the complex number in polar form. Express the answer in rectangular
form.
1. (1 + i) (1 - V3i). 2. (- 1 + i) (V + i).
3. (-1 + V3i) (\/ + 0. 4. l ~ l
fci'-VS.
1 + * V3 + 3f
7. -
.
V3 + i 1 + *
11
"'
2 " (2 + 1) (3 + 1)
13.,, .-.. .... 14.
(1 + ) (3 + 4i) (2 + 3t) (V3 + *)
.7.
[1 _ -1 100 / 1 1 _ \70
i(l - V2i)J 20. (-i + iV
In each of the problems from 22 to 29, find roots as directed and represent them
graphically.
22. Find two distinct square roots of 9 (cos 50 + i sin 50).
23. Find four distinct fourth roots of 16(cos 36 + i sin 36).
24. Find three distinct cube roots of 27(cos 165 + i sin 165).
25. Find five distinct fifth roots of - 32.
26. Find the three cube roots of 1.
27. Find the four fourth roots of 1.
28. Find the two square roots of i.
29. Find the three cube roots of - ^ (1 + V3i).
Sec. 11-8 Complex Numbers 203
In each of the problems from 30 to 34, the complex numbers E, 7, and Z designate
voltage, current, and impedance, respectively, and E = IZ.
30. Compute E when I = 5 + 4i amperes and Z = 30 - Si ohms.
31. Compute / when E = 110 + 3(K volts and Z = 20 - I5i ohms.
32. Compute Z when 7 = 4 + 3i amperes and E = 115 volts.
33. When two impedances Z\ and 2 are connected in parallel, the equation
--=+ determines an equivalent impedance Z. Compute Z when Z\
A ju\ L%
= 5 + 4f ohms and #2 = 8 6i ohms.
34. If 2 and z are conjugate complex numbers, prove that
12
Equations in
Quadratic Form
12-1. QUADRATIC EQUATIONS IN ONE UNKNOWN
This chapter provides an extension of the work on linear equa-
tions to second-degree, or quadratic, equations. Consider a quad-
ratic equation in one unknown written in the form
(12-1) ax 2 + bx + c = (a ^ 0),
where a, 6, and c are given real numbers. This equation is
called the general quadratic equation in x, and is said to be in
standard form.
If 6 T^ 0, (12-1) is called a complete quadratic equation; if b = 0,
it is called a pure quadratic equation. Thus, 3x~ a; + 4 = is a
complete quadratic equation in which a = 3, b 1, and c 4; and
x 2 2 = is a pure quadratic with a 1 and c 2.
In Section 12-4 we shall prove that every quadratic equation has
two and only two solutions or roots. The roots may be equal or
unequal, and they may be real or complex. Their natures depend
on the values of a, 6, and c. We shall consider the methods in gen-
eral use for finding these roots, and we shall then apply them as
well to the solution of equations which are not quadratic in x but
which can be written as quadratic equations in expressions involv-
ing the unknown.
12-2. SOLUTION OF QUADRATIC EQUATIONS BY FACTORING
If the left side of a quadratic equation in standard form can be
factored, the solution of the equation depends on the following
important principle :
The product of two or more numbers equals zero if and only if
at least one of the factors is equal to zero.
That is,
A B = if and only if A = or B = 0.
Sec. 122 Equations in Quadratic Form 205
In practice, we apply this principle by equating to zero each
linear factor of the left side of the given quadratic equation, and
solving the resulting linear equations. The following examples will
illustrate its application.
Example 12-1. Solve 2x* - 7x + 6 = by factoring.
Solution: To find the values of x which satisfy the equation 2x 2 7x + 6 = 0,
write the left side in the factored form
(x - 2) (2x - 3) = 0.
This product equals zero if and only if either
x - 2 = or 2s - 3 = 0.
Hence, x = 2 or x = 3/2. Moreover, 2 and 3/2 are solutions, because both 2 and
3/2 satisfy 2x* - 7x + 6 = 0. Thus,
2(2)2 _ 7(2) +6=8- 14 +6=0,
and
2(3/2) 2 - 7(3/2) +6=9/2- 21/2 +6=0.
Example 12-2. a) Solve the equation 2 sin 2 x - sin x - 1 = for sin x. b) Find
all non-negative angles x less than 360 which satisfy this equation.
Solution: a) Factor the given equation to obtain
(sin x - 1) (2 sin x + 1) = 0.
Since sin x 1 = or 2 sin x + 1 = 0, it follows that sin x 1 or sin # = 1/2.
Check:
2(1) 2 -(1) -1=2-1-1=0,
and
b) When sin x = 1, x = 90; when sin x = - 1/2, x = 210 or 330. Therefore,
x = 90 or 210 or 330.
Check: 2 sin 2 90 - sin 90 - 1 = 2 - 1 - 1 = 0,
2 sin 2 210 - sin 210 - 1 = 2( - ^) 2 - ( - ^) - 1 = 0,
and / 1\ 2 / 1\
2 sin 2 330 - sin 330 - 1 = 2^ - ^J - ( - ) - 1 = 0.
Note that this equation is a quadratic in which the unknown is a trigonometric
function of the angle x. We thus have only two values of sin x which satisfy the
equation. The determination of the angle, however, goes beyond the algebraic
solution of the quadratic, and it may happen that there are more than two values
of x which satisfy the equation. For this reason, it is recommended that all solutions
be checked by substituting in the original equation.
206 Equations In Quadratic Form Sec. 12-2
Example 12-3. Solve the equation 3 sec x = 2 cos x - 1 for all non-negative
values of x less than 27r radians.
1 3
Solution: Since sec x = - > we can write - = 2 cos x 1. We then clear
cos x cos x
of fractions, transpose, and factor, to obtain
2 cos 2 x - cos x - 3 = (2 cos j; - 3) (cos z + 1) = 0.
Hence, cos x = 3/2 or cos x = - 1. When cos x = - 1, x = TT. There is no
real number x for which cos x = 3/2.
Cfcecfc:
3 sec TT = 2 cos TT - 1, or 3(- 1) = 2(- 1) - 1.
It should be noted that factoring provides a method of solving any pure
quadratic equation ax 2 + c = 0. Thus, the equation is equivalent to
which gives
x + /i - =0 or x /i/ - = 0,
r d
or x = /!/ - This result agrees with that given by writing x 2 = -
r (Z &
and then simply extracting square roots of both sides to obtain
/ "~~
x = zb \/ - Note that the roots are real when - ^ and pure imaginary
^
when - > 0.
a
EXERCISE 12-1
Solve each of the following equations for x or 6. In each of the problems from 9
to 20, find all non-negative angles less than 360 which satisfy the given equation.
Check all solutions.
1. x 2 + 7x = 0. 2. z 2 - 10x +21 =0.
3. 2x(x + 5) = - 3. 4. z 2 + 6.r - 27 = 0.
5. x 2 - x = 6. 6. 6x 2 - 4x - 192 = 0.
7. o;3 + 27 - 3x(x +3) =0. 8. 16x - a 2 + 2a6 - 6 2 = 0.
9. sin 2 6 - sin = 0. 10. sin 6 = esc 0.
11. sin + 1 = 2 esc 0. 12. 2 tan 2 + 3 tan - 2 = 0.
13. sec (sec 6 + 6) = 16. 14. 3 cos 2 - 8 sin = 0.
2 esc 1 12 4
15. (cot - l).(cot + 2) = 4. 16.
3 6 esc 2 esc
17 cot + 4 143 1ft 8 2+3 cos _
cot + 6 ~ (cot + 6) 2 * 3 cos + 3 "*" 3 cos + 4 "^
19. 3 sin 2 0=4 sin* 0-10 sin 0. 20. A * ^ a " Q-^-2 - I = -
4 cos 2 8 cos 2
12-3. COMPLETING THE SQUARE
The method developed here is based on the fact that we can make
any binomial of the form x 2 + kx into a perfect square if we add to
Sec. 1 2-3 Equations in Quadratic Form 207
it the square of one-half the coefficient of x. To make this clear,
let us recall from Section 1-18 the formula for a perfect-square
trinomial. The formula is
(x + a) 2 = x 2 + 2ax + a 2 .
Since the coefficient of x in x 2 4- kx is k, the square of one-half of
(fc\ 2 A* 2
^J or -j Adding this to x 2 + fc#, we have
x 2 + kx + ^ = (x + ~)
Thus, the left member is a perfect square, namely, the square of
Applicability of the procedure to a variety of processes, including
solution of quadratic equations, is illustrated in the following
examples.
Example 12-4. Solve x 2 2x - 4 = by completing the square.
Solution: We first transpose the constant term, so that the left side will be of
the form x 2 + kx. Hence, the equation becomes
x 2 - 2x = 4.
Now the quantity ( I) 2 = 1 is added to the left side to make it a perfect square.
To obtain an equivalent equation, the same quantity is added to the right side also.
The result is
x 2 - 2x + 1 = 5,
or
(x - I) 2 = 5.
Taking square roots of both sides, we have
x - 1 = \/5.
So the desired solutions are x = 1 + \/5 and x = 1 V5.
Check:
j (1 + V5) 2 - 2(1 + V5) -4 = 1+2 \/5 +5-2-2 >/5 -4=0,
and _ _ _ _
(1 - \/5) 2 - 2(1 - V5) - 4 = 1 - 2 V5 + 5 - 2 + 2 \/5 - 4 = 0.
Example 12-5. Solve 2x 2 - 5x + 3 = by completing the square.
Solution: Transpose the constant term to obtain
2x 2 -5x = - 3.
Since the coefficient of x 2 is not 1, we make it 1 by dividing both sides by 2. Then
we have
o 5 3
208 Equations In Quadratic Form Sec. 12-3
(1 / 5\\ 2 25
2 \ / / = Tr "
2 5 . 25 _ 3 , 25 _
* ~ 2* + 16 ~ ~ 2 + 16 ~ 16
or
both sides, thus making the left side a perfect square. The result is
4.=-* +
^ 16 2
/ 5\ 2 _J_
\ X 4/ ~ 16 '
5 1
When we take square roots of both sides, we have x j = =t T Solving for z,
we obtain K 1
o l
X = -: -r'
4 4
That is,
= ~ or # = 1.
and
2(1) 2 -5(1) +3=2-5+3=0.
Example 12-6. Reduce x 2 + y* - 4x + Qy + 4 = to the form (x - ft) 2
+ (y - &) 2 = r 2 .
Solution: The solution of this problem requires that we complete the square of
the terms containing y as well as the square of those containing x. Hence, for
convenience, we write the equation in the form
(x - 4z ) + (y 2 + 6y ) = - 4.
When we complete the squares in the parentheses, the equation becomes
(x* - 4s + 4) + (0* + 6// + 9) = - 4 + 4 + 9.
Thus, the solution is
(x - 2) 2 + (y + 3) 2 = 9.
Example 12-7. Reduce 9z 2 - 4y 2 - l&c - 16# - 43 = to the form
A(x -ft) 2 - B(y -k)* =C.
Solution: Write the equation in the form 9(x 2 - 2x ) - 4(?/ 2 + 4t/ ) = 43.
Complete the squares in the parentheses to obtain
9(x 2 - 2s + 1) - 4(i/ 2 + 4y + 4) = 43 + 9 - 16.
Note that the numbers 1 and 4, which are added within the parentheses to complete
the squares, must be multiplied by the coefficients 9 and - 4, respectively, to
determine the numbers that are added to the right side.
The reduced form is, therefore,
9(s - I) 2 - 4(0 + 2) 2 = 36.
Example 12-8. Reduce \/3z 2 + 4z - 4 to the form y/a((x - h)* - A; 2 ).
Solution: For convenience, work with the quantity 3z 2 + 4# - 4 without the
radical sign until the final result is obtained. Hence, write
3* 2 + 4* - 4 =
Sec. 12-4 Equations in Quadratic Form 209
Complete the square of the terms in x and simplify to obtain
Now, write this result under the radical sign to obtain / l/3(( + ) <r)
Comparing this with the required form vX(# h) 2 /c 2 ), we see that we may
choose a = 3, h = - 2/3, and k = d= 4/3.
EXERCISE 12-2
In each of problems from 1 to 15, solve the given equation by completing the
square. In each of the problems from 9 to 15, find all non-negative angles less
than 300 which satisfy the given equation. Check all solutions.
1. x 9 - 8x = 20. 2. x 2 + lOx = 40. 3. x 2 - 7x = 30.
4. X 2 + x + i = o. 5. z 2 + x + 2 = 0. 6. 6x 2 - 5x - 1 = 0.
7. z 2 + z - 5 = 0. 8. 2x 2 = 3z + 9. 9. tan 2 = 2 tan + 1.
12. 1 + tan-' 6 = sec 0+3. 13. ^-4 = 2.
esc 2 01
14. 3 cot 2 + cot = 3 - 4 cot 0. 15. sec - cos = 2.
In each of the problems from 1(5 to 25, reduce the equation to the form
A(x -h) 2 +li(y -A-) 2 =C.
16. x 2 - 4?/ 2 - 2x + 1 = 0. 17. x 2 + 4i/ 2 - 6x + 16y + 21 = 0.
18. 4x 2 + % 2 + 32^ - 18?y + 37 = 0. 19. x 2 + 4// 2 - lOx - 40y -h 109 = 0.
20. 9x 2 + 4?y 2 - Sy - 32 = 0. 21. 4.c 2 -f 9^/ 2 - Ifxc - ISy -11=0.
22. x 2 - 9// 2 - 4x 4- 36;// - 41 = 0. 23. 4x 2 - 9// 2 + 32o; -h 36y -f 64 = 0.
24. 5// 2 - 4r 2 + 50# -f 32x + 41 = 0. 25. 3* 2 - y 2 + 2Qx - 2y + 11 = 0.
In each of the following problems, express the quantity inside the radical or
parentheses in the form a((x - h) 2 k 2 ).
26. V& ~ &r - 40. 27. v/2x - 16 J + 41. 28. (4// 2 - 20?/ - 76) 3 ' 2 .
29. * 30. (x 2 - 6.r 4- 34) 2/3 . 31. (2x 2 + 28.c + 34)~ 1/2 .
\/x 2 - Gx - 7
32. V(9^ 2 + 48,-c + 23) 3 . 33. (9z 2 + 24x + 25)~ 1/3 . 34. (7x* - 14a: + II)
12-4. SOLUTION OF QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA
By applying the method of completing the square to the general
quadratic equation (12-1), we can obtain a formula for the roots,,
either real or complex, of any quadratic equation whatever. The
general equation is
(12-1) ax 2 + bx + c = 0.
Transpose the constant term c, and obtain
ax 2 + bx = c.
2 1 Equations in Quadratic Form Sec. 1 2-4
Dividing by a, we have
2| 6 c
x * + x = --
a a
(1 6\ 2 & 2
- . - ) = ^ to both sides to obtain
2 o/ 4a 2
o . b . 6 2
* + :C+ - =
which becomes ft 2 &2 _ 4ac
Extracting square roots of both sides gives
, b __ Vb 2 - 4ac
* + 25~ 25
Solving for x, we have
6 -\/& 2 4ac
* = &
Hence, to solve a quadratic equation, put it into the standard
form ax 2 + 60? + c = 0, and substitute the coefficients a, &, and c in
the formula just derived to obtain the roots
,10 ON - & + Vb 2 - 4ac , - b - Vb 2 - 4ac
(12-2) xi = % and x 2 = ^
That these numbers #1 and # 2 are solutions of the given quad-
ratic equation is shown by substituting each of them in (12-1).
The details of this substitution for x l follow :
.
+c
/b 2 - 26 Vb 2 - 4ac + (6 2 - 4ac)
- -
b 2 + b Vb 2 - 4ac
2a
b 2 - 2ac - 6 Vb 2 - 4ac - 6 2 + 6 \b 2 - 4ac
- - .
Hence, the number x satisfies the equation ax 2 + bx + c = 0. A
similar computation shows that # 2 is also a solution of ax 2 + bx + c
= 0. Consequently, the two expressions given for x in (12-2) are
actually roots of (12-1). In Section 12-7 we shall study the expres-
sions in (12-2) further and shall determine when the roots are dis-
tinct and when they are real.
Sec. 1 2-4 Equations In Quadratic Form 2 1 1
Example 12-9. Solve 5z 2 6# 8 = by the quadratic formula.
Solution: Here a = 5, fr = 6, c = 8. Substituting these values in the
formula, we obtain
_
_ 6 db \/36 -f 160 6 db \/196 6 db 14
~ 10 10 10
Therefore, x\ = j^ = 2 and z 2 = -r = - g
These values are seen to satisfy the original equation when substituted for x.
Example 12-10. Solve x 2 - x + 2 = by the formula.
Solution: Since a = 1, 6 = 1, c = 2, we have
_ 1 Vr~^~8 _ 1 \^J _ 1 db
*~~ 2 - 2 "
Hence,
1+V? j 1
si = - ~ and z 2 =
Ctecfc:
Similarly, the second root may be checked.
Example 12-11. Solve 2 sin 2 x -f 3 cos x - 3 = by the formula, determining
all non-negative angles x less than 360.
Solution: We make use of the identity sin 2 x + cos 2 x = 1 to transform the
given equation into one involving a single trigonometric function of x.
Replacing sin 2 x by 1 cos 2 x t we have
2(1 - cos 2 x) + 3 cos x - 3 = 0.
Simplifying, we obtain
2 cos 2 x - 3 cos x -f 1 = 0.
Solving for cos x by the formula, we find
3 =t V9 - 8 3 1
cos x = - - A - = - A
4 4
Hence, cos x = 1 or 1/2. Therefore, x = or 60 or 300.
The solutions may be checked by substitution in the original equation.
Example 12-12. Solve cos x tan x -f sin 2 x = 1 sin x by the formula, deter-
mining all non-negative values of x less than 360.
Solution: Making use of the identity tan x = - > we have
6 - cos x
sin x , . .
cos x -- h sin 2 x = 1 sin x.
cos a:
2 1 2 Equations in Quadratic Form Sec. 1 2-4
Transposing and simplifying, we get
sin 2 x + 2 sin x 1 = 0.
Solving for sin x by the quadratic formula, we obtain
- 2 2
sin z =
- /o
= - 1 V2.
The value sin 3 = 1 + V2 is one solution. However, sin x = 1 \/2must
be excluded, since tho sine of an angle cannot be numerically greater than 1.
Check: For sin x = \/2 1, we find that
V2 -1
cos x = \/2 V2 - 2 and tan 3 = _
V2V2 - 2
Substituting these values in the original equation, we have
V2V2-2- ~ + (V2 - I) 2 = 1 - (V2 - 1).
- 2
This reduces to _ _
V2 - 1 + 3 - 2\/2 = 1 - V + 1
or
2 - x/2 = 2 - V2.
From the table of trigonometric functions, we have x = 2428' or 1 5532'.
EXERCISE 12-3
Solve each of the following equations for x or 8 by the quadratic formula. In each
of the problems from 14 to 24, find all non-negative angles 6 less than 360 which
satisfy the equation. Check all solutions.
1. s* + 6* - 7 = 0. 2. x 2 - x - 20 = 0.
3. x 2 + 2x - 5 = 0. 4. x 2 + x + I = 0.
5. z 2 + 2z + 1 = 0. 6. 7x2 _ gj. _ 9 = o.
7. 2z 2 4- 3* + 2 = 0. 8. 9.r 2 - 7x - 5 = 0.
9. (2x - I) 2 - 2(2x - 1) - 8 = 0. 10. (a 2 - 6 2 )z 2 - 4abx - (a 2 - 6 2 ) = 0.
5 - 6 8 - 3
11
11.
3+2 2X + 3 3 - 3 ' 1 - 3 3 + 1
13. - + * ; + o-Vr = 0- 14 - cot 2 + 3 = : 4
3 ^ x + 3 r 3x + 5 ~ v ' ww " ^ " " tan
15. -- = 3. 16. esc - 3 = sin 0.
tan cot
17. sin 2 - 2 sin = sin + 3. 18. 16 sec 2 - 8 sec + 1 = 0.
_ ft on cot * + 2 _ 2 cot 6 - 1 - o
~~
__ ___ _
2 ^ cos "*" 2 cos 2 2 cot - 3 cot
21. (sec 0+1) (sec + 2) = sec + 3. 22. (sin + 3) (3 - sin 0) = 3(sin + 3).
23. (esc + 2) 2 + esc = 1. 24. T + | = -
12-5. EQUATIONS INVOLVING RADICALS
Sometimes an equation in which the unknown appears under a
radical sign can be reduced to a quadratic by raising both sides to
Sec. 1 2-5 Equations in Quadratic Form 2 1 3
a power sufficient to remove the radical. The process must be
repeated until the unknown no longer occurs under a radical.
The operation of raising both sides of an equation to a power may
lead to an equation redundant with respect to the original ; that is,
the final equation may possess roots that are not roots of the orig-
inal equation. Such roots are called extraneous roots. For this
reason, every root obtained must be checked by substitution.
Example 12-13. Solve the equation \/x* - 3x + 4 = 2.
Solution: Cube both sides to obtain x 2 - 3x + 4 = 8. Transpose, and get
x 2 - 3x - 4 = 0.
Factoring and solving for x, we find that
x = 1 or x = 4.
Check:
and >X(-1) 2 -3(-l) +4 = ^1 +3 +4=^8=2,
_ 3(4) + 4 = ^/g = 2.
Hence, both 1 and 4 are roots.
Example 12-14. Solve the equation \/2x - 1 \/x +3 = 1.
Solution: Transpose one radical to obtain
- 1 = 1 + \A~T3.
When both sides are squared, the result is
2x - 1 = 1 + 2>A +3 + x + 3.
Combining like terms, we obtain _
x - 5 = 2 V-c + 3.
Now we square both sides to get
x 2 - Wx + 25 = 4(s + 3).
Transposing and combining gives
;r* - Ux + 13 = 0.
By factoring and solving, we find that
x = 1 or x = 13.
, V2(l) - 1 - VI +3 = 1 - 2 ?* 1,
and _ _
V2(13) - 1 - V13 +3=5-4 = 1.
Hence, 13 is a root, but 1 is not.
EXERCISE 12-4
Solve each of the following equations. In each case check for extraneous roots
1. V% - 2 = 4. 2. V3z +4 = 2.
3. Vz + 5 = 1. 4. ^3z -1=7.
5. \/z 2 - 16 = 2x l . 6. >A 2 - 2 = V2.c + 6.
214 Equations in Quadratic Form Sec. 12-5
7. x - 3 - V^T = 0. 8. * - 5* 1 ' 2 +6=0.
9. \/2x + 3 = 4 - 3x. 10. \/z - 1 + Vz - 3 = 2.
11. \/5 x 4- \X4oTT~5 = 5. 12. \/% x
12-6. EQUATIONS IN QUADRATIC FORM
Frequently, an equation which is not quadratic in the given
unknown may be considered as a quadratic in some expression
involving the unknown. Thus, or 4 - 3or 2 + 2 = and 2(x 2 - 2x) 2
(x 2 2x) 6 = may be treated as quadratic equations in the
expressions or 2 and (x 2 2x), respectively. This type of situation
was met earlier in Examples 12-11 and 12-12. The following
examples will further illustrate methods used in solving equations
in quadratic form.
Example 12-15. Solve the equation or 4 - 3or 2 +2=0.
Solution: Let x~ 2 = y, so that the given equation becomes
7/2 _ 3y + 2 = 0.
Factor, to obtain
(y - 1) (y - 2) = 0.
Therefore,
y = l or y = 2,
or
or 2 = 1 or z~ 2 = 2.
Hence, the solutions are
1
x = 1 and a: = 7=
Example 12-16. Solve (z 2 - 2) 2 - 7(.r 2 - 2) + 10 = 0.
Solution: Let x 2 2 = y, so that we get
y 2 - 7y + 10 = 0.
Factor, to obtain
(y - 2) (y - 5) = 0.
Hence,
y = 2 or y = 5,
or
x 2 - 2 = 2 or .T 2 - 2 = 5.
Then,
x 2 = 4 or z 2 = 7,
and the roots are
& = 2 and = \/7.
Sec. 127 Equations i'n Quadratic Form 215
Example 12-17. Solve x 2 + x - 2 V^ 2 + z + 3 = 0.
Solution: Let V# 2 + * + 3 = ?/. Then we can write
(x 2 + re + 3) - 2 -\A 2 + z + 3 - 3 = 0, or y 2 - 2y - 3 = 0.
Therefore, y = - 1 or 3, and \/z 2 + z+3=-lor Vz 2 + x + 3 = 3.
By definition of the radical, \/a is a non-negative number. Hence, although
\A 2 +05+3= 1 is consistent with the original equation, there are no values
of x which satisfy this equation.
Consider, then, V# 2 + # + 3 = 3. This leads to
z 2 + x + 3 = 9, or x 2 + a? - 6 = 0.
Hence,
x = 2 or a? = - 3.
Substitution shows that each of these values of x satisfies the original equation.
EXERCISE 12-5
Solve each of the following equations. Check all solutions.
1. ^ + X 2 _ 12 = 0. 2. 4jr* - liar* - 3 = 0.
3. (x 2 + 2) 2 + 3(z 2 + 2) - 4 = 0. 4. r* - 6z 2 + 8 = 0.
5. x 4 - 13x 2 + 36 = 0. 6. x 4 - 1 = 0.
7. (3* - 4) 2 + 6(3x - 4) + 13 = 0. 8. (x 2 + 3z) 2 - 14(z 2 + 3x) + 45 = 0.
9. f x + 1V + 2(2; + -) - 48 = 0. 10. (a; 2 - a:) 2 - 20(x 2 - x) + 36 = 0.
12-7. THE DISCRIMINANT
It will be recalled that the two roots of the general quadratic
equation ax- + bx + c = are
, - 4ac
and
2a
The expression 6 2 4ac, which appears under the radical sign, is
called the discriminant of the quadratic polynomial ax 2 + bx + c, or
the discriminant of equation (12-1).
In what follows we shall assume that a, 6, and c are real, and we
shall make use of the discriminant to determine the character of
the roots without actually solving the equation. By inspection of
the solutions Xi and x> 2 , we reach the following conclusions :
1. If 6 2 4ac = 0, each of the two roots #1 and x 2 is equal to
, and both roots are thus real.
2a
216 Equations in Quadratic Form Sec. 12-7
2. If 6 2 - 4ac is positive, then V& 2 ~~ 4ac is real, both roots are
real, and they are distinct.
3. If b 2 4ac is negative, then V& 2 ~~ 4ac is imaginary, and the
roots are distinct complex numbers of the form a + fti and a /3i.
These results may be summarized as follows :
Value of Discriminant
Character of Roots
>0
=
<0
Real and unequal
Real and equal
Unequal conjugate complex numbers
Furthermore, if a, 6, and c are rational, and ft- 4ac is a perfect
rational square, then the roots are rational; otherwise, they are
irrational.
The following examples will illustrate how to determine the
character of the roots.
Example 12-18. Determine the character of the roots of
2x* + 7x - 15 = 0.
Solution: Here a = 2, b = 7, c = 15. Hence,
62 - 4ac = 49 + 120 = 169 = (13) 2 .
The discriminant is positive, and so the roots are real and unequal. Since the
discriminant is a perfect square, the roots arc also rational.
Example 12-19. Determine all values of k for which the roots of the equation
kx 2 - 2kx -f 4 = are equal.
Solution: The discriminant must equal zero for the equation to have equal roots.
Hence, 6 2 - 4ac = 4& 2 - 16& = 4k(k - 4) = 0, and so k = or k - 4. When
k = 0, the equation is not quadratic. Therefore, the roots arc equal only when k = 4.
EXERCISE 12-6
Determine the character of the roots of each of the following equations by means
of the discriminant.
1. ar a + 3s + 4 = 0. 2. x 2 + &c - 9 = 0. 3. x* + IQx - 6 = 0.
4. 6z 2 - 7x + 3 = 0. 5. x 2 + IQx -f 2 = 0. 6. x 2 + 3x + 2 = 0.
7. 5z 2 + 7x + 2 = 0. 8. 4z 2 - 12z + 9 = 0. 9. x 2 - 2x + 1 = 0.
10. 4* 2 - 8z + 2 = 0. 11. 2x2 _ x + 3 = Q. 12. 5z 2 + z + 5 = 0.
13. 4x2 _ i 2x - 9 = o. 14. x 2 4 4z - 18 = 0. 15. x 2 - 4z + 7 = 0.
Sec. 12-8 Equations in Quadratic Form 217
12-8. SUM AND PRODUCT OF THE ROOTS
Adding the two roots of the general quadratic equation
ax 2 + bx + c = 0,
we obtain, by using (12-2),
- b + \/b 2 - 4ac , - 6 - V b 2 - 4ac 26 6
xi+x*= 2 - + Ya = -_=--.
Also, multiplying these roots, we have
- 6 + Vb 2 -ac - b - Vb 2 - 4ac
XlX2 = 2 -
__ b 2 (b 2 4ac) __ 4ac __ c
"" 4a 2 "" 4a 2 "" a
Hence, we have, for the sum and product of the roots,
(12-3) xi + x 2 = ~ - >
and
(12-4) xix a =-
tZ
These formulas are used in various ways, for example, in check-
ing roots of a quadratic equation, and in forming an equation if its
roots are known.
To find the factored form of the quadratic polynomial ax 2 + bx
+ c, let us make use of the sum and product formulas just found.
Solving Xi + x.> = for 6, and solving XiX 2 = - for c, we have
a a
b = a(xi + x< 2 ) and c a(x^x^). Hence, by substitution, we have
ax 2 + bx + c = ax 2 a(x\ + x<)x + a(x\X2)
= a(x 2 - (xi + x 2 )x + (ziofe))
= a(x - xi) (x - X2).
Therefore, if x\ and x> 2 are the roots of the quadratic equation
ax 2 + bx -f c = 0, its factored form can be written
a(x xi) (x #2) = 0.
Since a ^ 0, we may write the equation equivalently as
(X - Xi) (X 3fe) = 0.
This form or its expansion,
X 2 (#1 + X 2 )X + XiX2 = 0,
may be used in writing an equation whose roots are known.
Example 12-21 indicates the procedure.
Example 12-20. Without solving the equation, find the sum and the product
of the roots of the equation 3x 2 5.c +2 =0.
Solution: In this case, a = 3, b - 5, and c = 2. Then the sum of the roots is
= - , and the product of the roots is - = ^
ft o do
218
Equations in Quacfrof/c Form
Sec. 12-8
Example 12-21. Write a quadratic equation in the form (12-1), given that the
roots are (1 \/3 i).
Solution: Since x\ = 1 + \/3 i and x 2 = 1 V3 t, we have
and Xl + X2 = (1 + ^ l>) + (1 ~~ ^ f) = 2 '
sia* = (1 + \/3 i) (1 - V3 i) = 1 - 3 i a = 4.
Therefore a suitable equation is x 2 2r -f 4 = 0.
Alternate Solution: Writing the desired equation in factored form, we have
(* - (1 + VSO) (*-(!- V3 <)) = 0.
Simplifying, we obtain
(z - 1 - \/3 i) (x - 1 + V3 i) = 0,
((* - 1) - V3 i) ((* - 1) + x/3 f) = 0.
(3 - 1) + 3 = 0,
a; 2 - 2x + 4 = 0.
or
Hence,
and, finally,
EXERCISE 12-7
In each of the problems from 1 to 9, find the sum and the product of the roots
of the given equation.
1. z* + 2* - 1 = 0. 2. 3.r 2 - x + 2 = 0. 3. a* + 2 = 0.
5 -ir-4*+i= -
6. 5.D 2 - 6z + 1 = 0.
4. 6x 2 - 2.c + 3 = 0.
7. 5x* - Qx - 1 = 0. 8. 5 2 4- 6.c + 1 = 0. 9. lOOx 2 - 40x +17-0.
Form an equation with each of the following pairs of roots.
13. 3, 6.
10. 1, - 3.
1/t 3 1
14 '2'3'
11. 0, + 2.
15. 2 iV
18. V3, V5. 19. 0, V3 -
12. - 1, 1.
16. |(1
20. a 6f .
17. i.
21. V3
12-9. GRAPHS OF QUADRATIC FUNCTIONS
To graph any quadratic function of
the form ax- + bx + c, we set y ax 2
4- bx + c and construct a table of values
of y corresponding to assigned values of
x. The graph is of the type shown in
Fig. 12-1 and is called a parabola.
As found by the quadratic formula,
the two solutions of the general quad-
ratic equation (12-1) are given by
(12-2)
and
Xl =
4ac
2a
FIG. 12-1.
6 \/b 2 4ac
X2 =
Sec. 1 2-9 Equations in Quadratic Form 2 1 9
Since (12-1) states that y = in the equation y = ax 2 + bx + c, the
solutions Xi and x 2 are ^-intercepts of the curve.
In Fig. 12-1, let A and C be the ^-intercept points, and let B be
the mid-point of AC. We note that
OB = OA + AB,
or . 7
- Xl "" X2 - Xl +X2 -
Therefore, B is the point ( ^- > V
Now consider the equation y = fc, which represents a straight
line parallel to the x-axis. This line may or may not intersect the
parabola, depending on the value of k. Solving the equations y = k
and y = ax- + bx -f c simultaneously, by elimination of ?/ we obtain
ax- + bx + c k = Q. The roots of this resulting equation are
and
6 \/6 2 4.a (c /c)
2a 2a
If the value of k is such that y k intersects the curve in two
distinct points, the discriminant in (12-5) is greater than zero,
and the roots will be real and distinct. The point on the line y k
with abscissa x = - ^- is then equidistant from the points of inter-
ci
section, whose abscissas are x and x z .
If, on the other hand, the value of k is such that y = k does not
intersect the curve, the discriminant in (12-5) is less than zero,
and we have a pair of conjugate complex roots, In this case,
z>a
is the real part of these roots.
The points with abscissa x = and arbitrary ordinates lie
0.
on a vertical line, called the axis of symmetry, or simply the axis
of the curve; the curve is said to be symmetric with respect to
the axis.
The point of intersection of the axis and the parabola is called
the vertex of the parabola. If the coefficient a of the second-degree
term of y = ax- + bx + c is positive, the vertex is the loivest point,
and the curve is said to be concave upward. If a is negative, the
220
Equations in Quadratic Form
Sec. 12-9
vertex is the highest point, and the curve is said to be concave
doivmvard.
To find the coordinates of the vertex, we solve simultaneously
the equation x = of the axis and the equation y = ax 2 + bx + c
of the parabola. The coordinates of the vertex are thus found to be
b , a& 2 b 2 , b 2 -
(12-6)
and y = -r-o - -^
4o
The vertex may be characterized in another way. Let us demand
that a horizontal line y = k intersect the parabola in two coincident
points. That is, let us insist that the two roots x\ and x s in (12-5)
coincide. Then the discriminant in (12-5) is equal to 0, and
#1 = x 2 = ^-- The value of y corresponding to this value of x is
Za
the ordinate of the vertex, as found in (12-6) . We say that the line
b 2 - 4ac
y = -
4a
is tangent to the parabola at the vertex.
Example 12-22. Graph x 2 - Gx + 4.
Solution: Let y = x 2 - 6x -f 4, assign values to x, and compute the corres-
ponding values of ?/, as in the accompanying table. The graph is shown in Fig. 12-2.
The coordinates of the vertex are
(-i.il)
=!-
(7,11)
b 2
and 2/=c ^-=4 9= -5.
(2, -4)
X
y
- 1
11
4
1
- 1
2
-4
3
-5
4
-4
5
- 1
6
4
7
11
Hence, the axis is the line x = 3. Since a is
positive, the vertex (3, 5) is the lowest point
on the curve, and the curve is concave upward.
Sec. 12-10
Equations in Quadratic Form
221
Example 12-23. Graph y = x 2 - Qx + 9 and ?/ = s 2 - Qx + 14 relative to the
same coordinate system as was used for the graph of y = x 2 6# -f 4.
Solution: Tables similar to that in Example 12-22 but applying to the first two
curves are constructed. The three curves are shown in Fig. 12-3.
(C)
X
y = X 2 - ftp + 9
- 1
16
9
1
4
2
1
3
4
1
5
4
6
9
7
16
y - j-2 _ Ox + 14
- 1
21
14
1
9
2
6
3
5
4
6
5
9
6
14
7
21
FIG. 12-3.
Curve (A) crosses the x-axis at two points, corresponding to the roots 3 =fc \/5
of .r 2 Gx H- 4 0. Curve (#) is tangent to the x-axis at (3, 0), because both
roots of x 2 6.c + 9 = an 1 equal to 3. Curve (C) does not intersect the z-axis,
because 1 x 2 Gx + 14 = has imaginary roots.
The reader should relate the discriminants of the quadratics to a study of these
graphs.
12-10. QUADRATIC EQUATIONS IN TWO UNKNOWNS
The general equation of the second degree in x and y is
(12-7) ax 2 + bxy + cy 2 + dx + ey + f = 0,
where a, b, c, d, e, and / are given real numbers. An equation of
this form, in which at least one of the coefficients a, 6, and c is
different from zero, is called a quadratic equation in x and y.
By a solution of such an equation, we mean a pair of real or
complex numbers which, when substituted for x and y in (12-7),
will reduce the left side of the equation to zero. Usually there are
infinitely many pairs of numbers which satisfy the equation.
222 Equations in Quadratic Form Sec. 12-10
If c T^ 0, (12-7) may be solved for y in terms of x by means of
the quadratic formula. Corresponding to each real value assigned
to x f we then obtain, in general, two values of y. We then have
pairs of numbers (x, y) which, if real, may be plotted in a rec-
tangular-coordinate system. It is shown in analytic geometry that
the graph so obtained will be one of a class of curves called conic
sections, inasmuch as they may be obtained as curves of intersec-
tion of a plane and a right circular cone. At this time we shall
confine ourselves to merely listing the curves which comprise this
class and indicating briefly the form of the quadratic that corre-
sponds to each of the graphs. A more adequate discussion of this
subject is given in analytic geometry. However, typical examples
of these curves are given here.
1. Parabola. When A ^ 0, the equations y = Ax- + Bx + C and
x = Ay- 4- By + C represent parabolas with vertical and horizontal
axes of symmetry, respectively.
2. Circle. When C is positive, the equation x-+ y 2 = C represents
a circle whose center is at the origin and whose radius is \A?-
3a. Ellipse. When the constants are positive, the equation
Ax- + By- = C represents a curve called an ellipse. If A = B, the
ellipse is a circle.
3b. Point Ellipse. If A and B are positive and C 0, the equa-
tion Ax- + By 2 = C is satisfied by only one point, namely, the origin.
The graph is then said to be a point ellipse.
3c. Imaginary Ellipse. If A and B are positive and C < 0, there
are no (real) points on the graph, and we say that the equation
Ax 2 + By 2 C represents an imaginary ellipse.
4a. Hyperbola. When A, B, and C are positive, the equations
Ax- By 2 = C and Ay 2 Bx 2 = C represent hyperbolas.
4b. Hyperbola. When C ^ 0, the equation xy = C represents a
curve called an equilateral hyperbola.
5. Pair of Straight Lines. The equation Ax 2 + Bxy + Cy~ + Dx
f Ey + F = represents two straight lines, which may be either
distinct or coincident, if the left side can be expressed as the prod-
uct of two real linear factors.
We use the quantity b 2 4ac, which is usually called the charac-
teristic of the equation ax 2 -f- bxy + cy 2 -f dx + ey -f / = 0, to deter-
mine the nature of the conic corresponding to a particular form of
the general quadratic equation. In analytic geometry the following
statements are shown to be true :
1. If b 2 4ac = 0, the conic is a parabola or two real or imagi-
nary parallel lines.
2. If b' 2 4ac < 0, the conic is an ellipse or a point ellipse or an
imaginary ellipse.
Sec. 12-10
Equations In Quadratic Form
223
3. If b 2 4ac > 0, the conic is a hyperbola or two intersecting
lines.
In Section 12-9 the graph of the parabola y = ax 2 + bx + c was
discussed. In the following illustrative examples, the procedures
for graphs of other quadratic equations are considered.
Example 12-24. Graph x 2 + y 2 = 9.
Solution: Set y = to obtain the ^-intercepts, which are 3; and set x =
to obtain the ^-intercepts, which are it 3. Solve the equation for y, obtaining
y = V9 - x 2 .
Then construct a table of other corresponding values pf x and y. To yield real
values of y, the numerical value of x cannot exceed 3. As shown in Fig. 12-4, the
resulting graph is a circle with center at the origin and radius 3.
IY
X
y
-4
-3
imaginary
-2
2.24
- 1
2.83
3
1
2.83
2
2.24
3
4
imaginary
-3
FIG. 12-4.
Example 12-25. Graph 4z 2 + 9?/ 2 = 36.
Solution: Set y = to obtain the ^-intercepts, which are 3; and set x = to
obtain the ^-intercepts, which are 2. Solve for y and obtain
y = V9 - x 2 .
o
Construct a table and draw the curve, as shown in Fig. 12-5. This illustrates
an ellipse.
X
-3
- 2 1.49
- 1 1.89
2
1.89
1.49
FIG. 12-5.
Note that the numerical value of x must be equal to or less than 3 in order to
yield real values of y.
224 Equations in Quadratic Form Sec. 12-10
Example 12-26. Graph 4z 2 - 9s/ 2 = 36.
Solution: Setting y = 0, we find that the ^-intercepts are 3. Setting x = 0,
however, results in the equation y 2 = - 4. Hence the curve has no ^/-intercepts.
Solving the given equation for ?/, we have
y = | V* 2 - 9-
The accompanying table is constructed.
a:
y
-6
3.5
-5
2.7
-4
rfc 1.8
-3
-2
imaginary
2
imaginary
3
4
1.8
5
2.7
6
3.5
FIG. 12-6.
Note that in this case the numerical value of x must be equal to or greater than 3 in
order to give real values of y. The graph of the given equation is shown in Fig. 12-6.
This illustrates a hyperbola.
EXERCISE 12-8
Identify and graph each of the following.
+ 9z 2 = 36. 3. 4,r 2 - 9?/ 2 = 0.
- 4*/ 2 = 16. 6. x 2 + 9?y 2 = 0.
= - 4. 9. 5^ 2 -f- ftcy = 28?/ 2 .
11. 5ar?y = 2x + ?/.
13. x 2 + xy - 2y* + 3y - 1 = 0.
2x + 4?y - 12 = 0.
16. z 2 - 3x - 3y* + I8y - 27.
18. 4x 2 + 4xy - 3i/ 2 + 4x + lOy = 3.
20. 4x 2 + 3xy + 4i/ 2 - Sx - 8y = 24.
12-11. GRAPHICAL SOLUTIONS OF SYSTEMS OF EQUATIONS INVOLVING
QUADRATICS
In Chapter 9 we solved systems of two or more linear equations
both algebraically and graphically. Frequently, however, simul-
taneous systems include one or more equations of the second or
higher degree. We have, therefore, to consider the problem of find-
ing systems of values of the unknowns x and y that satisfy two equa-
1. x 2 + y 2 = 25. 2.
4. 4z 2 - 9*/ 2 = 36. 5.
7. y = x 2 - 3x + 2. 8.
10. 2z 2 + ?/ 2 - 4y =4.
12. 3z 2 - 4xy + 2^/ 2 - 6x + 3?/ = 7.
14. 4* 2 - 4ry + ?y 2
15. xy + 2/ 2 - y - 2x - 2 = 0.
17. 2x 2 - xy - 28y 2 = 0.
19. 9z 2 - 24xy + 162/ 2 + 3x - 4?/ = 6.
Sec. 12-11
Equations in Quadratic Form
225
tions, one of which is quadratic and the other of which is linear or
quadratic.
We shall begin by illustrating some graphical solutions of several
types of systems. The graphical method yields only the real solu-
tions of a system, but it may prove advantageous in suggesting
solutions and interpreting results. In general, this method yields
at best only approximate solutions. The graphs should be drawn as
accurately as possible.
Example 12-27. Solve graphically the system
' x 2 - 8* + 3y = 0, *
x - Zy + 6 = 0.
Solution: Solving each equation for y in terms of x, we have
8x - x 2 , x + 6
y= and =-3
Construct tables of values, and draw both graphs, using the same coordinate
system, as shown in Fig. 12-7.
x
8x -x 2
V 3
- 1
-3
1
7/3
2
4
3
5
4
16/3
5
5
6
4
7
7/3
8
9
-3
x + 6
X
Q
-6
2
7
123456 7 8\
(A): J?2 -8.1: + 3^-0*
(B): *-
+*x
FIG. 12-7.
The line and the parabola are seen to intersect at the points (1, 7/3) and (6, 4).
It follows that these points represent common roal solutions, possibly only approxi-
mate, of the system. A check by substitution shows that the real solutions are,
in fact, x = 1, y = 7/3; and x = 6, y = 4.
The solution of Example 12-27 suggests a procedure for finding
graphical solutions of quadratic equations in one unknown.
226
Equations in Quadratic Form
Sec. 12-11
Example 12-28. Solve the equation x* - x 2 = graphically.
Solution: Since x 2 = x + 2, both sides of this equation may be set equal to y.
Thus, the original equation is replaced by the system
f y = x\
( y=x+2.
y = x + 2
d=l
2
3
4
1
4
9
16
-2
2
From the accompanying tables of values of x and ?/, the graphs shown in Fig.
12-8 are constructed. The graphs show that the line and the parabola intersect at
the points ( 1, 1) and (2, 4). Since these points are common to both graphs,
their abscissas must satisfy the equation
x 2 = x + 2.
Hence, the required roots of the original equation x 2 x 2 = are x = 1
and x = 2.
'(-2,0)
FIG. 12-8.
Example 12-29. Solve graphically the system
( t
3x 2 - 2?/ 2 = 6.
Solution: From the first given equation,
and
From the second equation,
and
Sec. 12-12 Equations in Quadratic Form 227
The necessary tables are given here, and the graphs are shown in Fig. 12-9.
X
1
x
. /3 X 2 _ 6
/ii i ^ /I A 2
y =t 2 v 10 x^
y _L /j/ 2
2
imaginary
1
^V15
1
imaginary
2
V3
v/2
3
-*V7
2
V3
4
3
V1^5
- 1
- V15
4
=b V21
2
-2
V3
-2
db V3
-3
\V7
-3
V1^5
-4
-4
1/21
The ellipse (A) and the hyperbola (B) are seen to intersect in the following four
distinct points:
(2, V3), (2, - x/3), ( - 2, V3), ( - 2, - >/3).
These values of x and y already appear in the tables used for constructing the
graphs, and need not be checked by substitution in the original equations. However,
such checking is usually desirable.
EXERCISE 12-9
Solve each of the following systems of equations graphically.
1.
4.
7.
10.
13.
12-12. ALGEBRAIC SOLUTIONS OF SYSTEMS INVOLVING QUADRATICS
As in the case with linear equations discussed in Section 9-1, it
may happen that in a system of equations involving quadratics
'x - 2y +3=0,
2.
x + ?/ = 4
3.
1
'x - y + 1 = 0,
,r 2 = 3?/.
?y 2 = 2x.
x- 2 + ?y 2 = 25.
y* = 3,
5.
'x+y= 6,
6.
3,r + 2y = 6,
3x + y = 6.
.r 2 = y.
xy = - 12.
2x -y =4,
8.
1
x 2 - 7/ 2 = 16,
9.
82 + 3y = 25,
xy = 6.
x + 3j/ = 4.
4z 2 + 2/2 = 25.
> - 2v/ 2 - 4 = 0,
11.
"x 2 + y 2 = 13,
12.
9# 2 + 4y 2 = 36,
.r 2 -9^=0.
x 2 = 122/.
z 2 +2/2 =81<
'9.r 2 +25?/ 2 =225,
14.
,r 2 + z/ 2 = 20,
15.
a:2 + y * = 20,
^2+2/2- 4.
7/2 _ X 2 = 12.
4^2 + 9^2 = joo; 1
,
16.
1
* + y = o,
z2 4- 1/ 2 = 8.
228 Equations in Quadratic Form Sec. 12-12
part of the graph of one equation coincides with part of the graph
of the other. Such a condition gives rise to infinitely many solu-
tions. Usually, however, there are only a finite number of points
of intersection of the graphs corresponding to the given equations,
and the algebraic problem consists of finding the pairs of numbers
(x, y) which satisfy both equations. We can say in this case that
two simultaneous equations in x and ?/, of degrees m and n, respec-
tively, can have at most mn solutions. Thus, a system of one linear
and one quadratic equation can have at most two solutions, and a
system of two quadratics can have at most four solutions.
When a system consists of two quadratic equations, the algebraic
solution usually leads to a fourth-degree equation in one of the
unknowns. Since we have not presented a general method of solving
a fourth-degree equation, we shall consider here only systems whose
solutions can be effected by the theory of quadratic equations. The
methods of procedure in some of the more important types are
shown in the following three cases :
Case 1. One Linear and One Quadratic Equation. A system of
this type can always be solved by the method of elimination by
substitution.
Example 12-30. Solve the system
x - 3?/ + 6 = 0,
x 2 - Sx + Zy = 0.
Solution: Solve the linear equation for y in terms of x, obtaining
x + 6
Substitution for y in the quadratic equation yields
,,_ 8a;+ 3(?L6)=0.
Collecting terms gives
x 2 - Ix + 6 = 0.
The roots of this equation are x = 1 and x = 6. Substituting these values in the
linear equation, we obtain y = 7/3 and y = 4. Hence, the solutions are
x = 1, y = 7/3; and x = 6, y = 4.
These values can readily be verified as solutions of the given system. Note that
they correspond to the coordinates of the points of intersection in Fig. 12-7.
Example 12-31. Solve the system
x + 2y + 4 = 0,
x* + 4y* - 2x - 3 = 0.
Sec. 12-12 Equations In Quadratic Form 229
Solution: Solve the linear equation for 2y, to obtain
2y = - (x + 4).
Substitute in the quadratic and collect terms. We then have
* z 2 + (x + 4)2 - 2z - 3 = 0,
or
2x 2 + 6x + 13 = 0.
One solution is
- 3 + i V17 5 + t \/17
*= 2 ' y= 4
The other solution is
- 3 - tVl7 5 - i yT7
* = 2 ' y= '
Since these values arc imaginary, the graphs of the two given equations do not
intersect.
Case 2. Two Equations of the Form ax 2 + by 2 = c. When the
system consists of two equations containing only squared terms in
each unknown, it can be solved for x 2 and y 2 by the methods used
for linear systems in Section 9-2.
Example 12-32. Solve the system
x* + 2y* = 17,
2x 2 - 2/ 2 = 14.
Solution: To eliminate y 2 , multiply the second equation by 2 and add the two
equations, as follows:
x 2 + 2y* = 17
4z 2 - 2y 2 = 28
5.r 2 = 45.
Solving for x, we have
i = 3.
Now substitute 9 for x 2 in the first of the original equations. Then
2?/ 2 = 17 - 9 = 8,
or
y = 2.
Hence, we have the following four solutions:
(3,2), (3, -2), (-3,2), (-3, -2).
These may be written (3, =b 2), ( - 3, d= 2).
Case 3. Two Equations of the Form ax 2 + bxy + cy 2 = d. If the
system is of this type, the solution is effected by elimination of the
constant term. The resulting equation is then solved for one
unknown in terms of the other. This procedure gives us two linear
equations in x and y which may be combined with either of the
given quadratic equations to form two systems of the type con-
sidered in case 1.
230 Equations in Quadratic Form Sec. 12-12
Example 12-33. Solve the system
x* - xy + 2y 2 = 1,
2x 2 _ 2xy + Sy 2 = 3.
Solution: Multiply the first equation by 3 and subtract the second given equation
from the new equation, as follows:
3z 2 - 3xy + 6?/ 2 = 3
2x 2 - 2xy + Sy 2 = 3
x 2 xy - 2y 2 = 0.
Factor, to obtain
(x + y) (x - 2y) = 0.
Hence,
x + y = or x - 2y = 0.
We may combine each of these two equations with the first given equation to
form the following two systems:
x + y = 0;
and
f x 2 - xy + 2y 2 = 1,
f x 2 - xy + 2y 2 = 1,
- 2y = 0.
We then proceed by the method for case 1.
The solutions of the given system consist of the solutions of these two systems.
Hence, we have
x = 1/2, y = - 1/2; s = - 1/2, y = 1/2;
* = l,y = l/2; *= -1,0 = -1/2.
Occasionally, another method is effective in connection with systems described
under case 3. The following example illustrates this method.
Example 12-34. Solve the system
f x 2 + 9?/ 2 = 37,
I xy=2.
Solution: Multiply the second equation by 6, to obtain 6xy = 12. Add this to*
the first equation, to obtain
x 2 + 6xy + 9y 2 = 49.
Also subtract Qxy = 12 from the first equation to obtain
x 2 - Qxy + 9?/ 2 = 25.
The left side of each of these new equations is a perfect square. We chose the
multiplier of xy, which is 6 in this case, so as to obtain perfect squares. We now
have the system
/ (x + 3y) 2 = 49,
I (x - i
3y) 2 = 25.
Hence,
x + 3y = 7 or x + 3y = - 7.
Also,
x 3y = 5 or x 3?y = 5.
We now solve the following four systems of linear equations:
+ 3y = 7, fa; + 3y = 7, fa? + 3y = ~ 7, f 3 +
-3y = 5; \a? - 3y = - 5; \a?-3y=5; \x -
Sec. 12-13
Equations in Quadratic Form
231
An equation in x and y is said to be symmetric in x and y if the
equation is unchanged when x and y are interchanged. When a
system consists of two quadratic equations both of which are sym-
metric in x and y, the substitutions x = u + v, y-u v will give
an equivalent system which may in some cases be solved by previ-
ous methods.
EXERCISE 12-10
In each of the problems from 1 to 21, solve the given system of equations
algebraically.
1.
x -2y +3 =0,
2.
x + = 4,
3.
x - y + 1 = 0,
x* = 30.
?/ 2 = 23.
x 2 + 2 = 25.
4.
V = 3x,
5.
3z +20 =6,
6.
'23 - = 4,
33 + y = 6.
x?^/ + 12 = 0.
x0 =6.
7.
x 2 - y 2 = 16,
8.
8x + 30 = 25,
9.
0-3=2,
x + 3y = 4.
4z 2 + 2 = 25.
x 2 + y* - 2x - 40 = 20.
10.
'x 2 - 2?/ 2 = 4,
11.
x 2 + ?/ 2 = 10,
12.
(x - 3)2 + (0 - 1)2 = 16,
X 2 _ g ; , y = o.
z 2 = 90.
(x - I) 2 + (y - 1)2 = 12.
13.
9x 2 + 4?/ 2 =36,
14.
9x 2 + 25?/ 2 = 225,
15.
x* + y 2 = 20,
3*+0'=81.
x 2 + ?/ 2 = 4.
7/2 _ X 2 - 12.
16. jxy = 2,
17 - (i + 1 - ? ,
18. fx 2 + 3x0 = 10,
\ X 2 _y2 = 3.
|z 2 '
1x0 = 3.
(x + 77 = 3.
19. (x* + 330 = 28,
20. |3.r 2 +7/ 2 =28,
21. [30 + 02 = 12,
[30 + 4?/ 2 = 8.
\4x 2 -30 +0 2 =40.
jx0 = 2x 2 - 24.
22. Complete the solution of the system in Example 12-34.
In each of the problems from 23 to 26, solve the given system by the method of
Example 12-34.
23.
25.
x 2 + 7/2 = 50,
xy = 25.
X 2 + V 2 - 25,
xy = 12.
24.
26.
z 2 + 42/2 = 13,
xy = 3.
*2 + 7/2 = 144,
a?y = 56.
Solve each of the following symmetric systems.
27.
29.
X 2 + y2 = 4;
xy + x + y = 10.
z 2 /?/ + 2 A = 56,
3+0=2.
28.
30.
x 2 + 02 - x - = 2,
xy + 3x + 30 = 2.
x 2 + 02 = i 3>
3x 2 + 2x0 + 30 2 = 42.
12-13. EXPONENTIAL AND LOGARITHMIC EQUATIONS
An equation in which the unknown occurs in an exponent is
called an exponential equation. Such an equation is usually solved
by taking the logarithm of each side and solving the resulting
equation. When this latter equation is in linear or quadratic form,
it may be solved by preceding methods.
232 Equations in Quadratic Form Sec. 12-13
Example 12-35. Solve for x: 5* +3 = 625.
Solution: Write the equation in the form
5*+ 3 = 5*.
This equation is satisfied if and only if x + 3 = 4, that is, x = 1.
Example 12-36. Solve the equation 2 3 ** 1 = 3 4 *.
Solution: Taking the logarithm of each side to the base 10, we get
log (2**+*) = log (3 4 *),
or
(3z + 1) log 2 = 4z log 3.
Therefore,
_ _
4 log 3 - 3 log 2
From Table III, log 2 = 0.3010 and log 3 = 0.4771. Substituting these values,
we have
_ 0.3010 __
X ~ 4 0.4771 - 3 0.3010 '
or
Example 12-37. Solve for x: log (x + 1) = 1 - log (3a? + 2).
Solution: Collecting terms containing logarithms on one side and writing that
member as a single logarithm, we have
log (x + 1) + log (3x + 2) = 1,
or
log (x + 1) (3x + 2) = 1.
Writing the members in exponential form, we have
(x + 1) (3* + 2) = 10' = 10.
This equation reduces to
3z 2 + 5z - 8 = 0.
g
Solving for x, we find that a; = 1 or a; = '
o
o
Checking, we find that x = 1 satisfies the original equation, whereas x = -
o
gives rise to logarithms of negative numbers. Since negative numbers do not have
real logarithms, this latter value of x is not to be used.
Example 12-38. Solve y = log, (x + \/l + # 2 ) for x in terms of y.
Solution: Write the given equation in exponential form, obtaining
Transpose and square to remove the radical. The result is
e*v - 2xe -1=0.
Solving this equation for x, we have
Sec. 12-14 Equations in Quadratic Form
EXERCISE 12-11
Solve each of the following equations for the unknown x.
233
1. 2* = 64.
4. 10* = 0.0001.
7. 5* = 15.
10. (3*) (2*) = 36.
13. log* 2 = 0.6932.
16. 5.03 = (3.17)i'<*-i>.
19. log (3x - 5) = 3 - log 7.
21. z 10 *** 3 ) = 1000.
23. e 2 * = 4.83.
26. e* 2 + 2 *- 2 = 16.
29.,=^.
2. 4** 1 = 256.
5. 4* = 24.
8. 3* = 17.
11. 5"* = 2.403.
14. s 3 l4 =0.04681.
17. 3*+ 2 =
3. 3 3 ** 1 = 243.
9. 3(2*) = 6*.
12. log* 8 = 0.4136.
15. (1.5)* = 32.
18. Iog 2 x = 3.
20. log (4* - 1) t= 1 - log (to + 2).
22. Iog 2 (-!)+ Iog 2 (x + 3) = 3.
25. 4e*+* = 7.
31. e 4 * - e 2 * - 10 = 0.
12-14. GRAPHS OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
The graph ofy = log x is shown in Fig. 12-10. By assigning
values to x, one finds corresponding values of y from Table III. A
few pairs of values are shown in the accompanying tabulation.
X
y
0.1
- 1
0.2
-0.70
0.3
-0.52
0.5
-0.30
1
2
0.30
3
0.48
4
0.60
10
1.
FIG. 12-10.
The graph of # = e* may be obtained from a table of exponential
functions. Here, however, we shall proceed as follows: Take the
logarithm of each side to the base 10, to obtain
log y = x log e.
234
Equations in Quadratic Form
Sec. 12-14
Prepare the accompanying table, and construct the graph in Fig.
12-11.
X
log e
x log e
y
0.434
1.0
1
0.434
0.434
2.7
2
0.434
0.868
7.4
3
0.434
1.302
20.0
4
0.434
1.736
54.5 .
Graph each of the following.
1. y = Iog 7 x.
*-"*-?
4.y = 10-* 2 .
7. y =3+4(2-).
10. y = 100e- 06 *.
EXERCISE 12-12
2.x = Iog 7 y.
2
x
5. y = e 2 .
8. y = e"'.
11. v = 7 3 *- 1 .
3. T/ = 3-.
6. y = 16(5-).
9. y = log c x.
12. ?/ = 3.9 2 *- 3 .
13
Theory of Equations
13-1. INTRODUCTORY REMARKS
With the ever-increasing importance of mathematics in engi-
neering and the physical sciences, problems are constantly occur-
ring that involve the solution of equations. Often these equations
are of the simple algebraic or trigonometric types which we have
already learned to solve. There are many other problems, however,
which require the solution of equations of higher degree than the
second and of some types of somewhat more complicated tran-
scendental equations. Equations of the third and fourth degree can
be solved by methods analogous to those which we used for quad-
ratic equations. Because of their complexity, however, these
methods are seldom used. It has been proved that no such proce-
dures exist for equations of degree higher than the fourth.
In this chapter we shall consider various properties of poly-
nomial equations in general. Some of these properties will be of
considerable use in later studies of mathematics, while others are
considered here merely for the aid they give us in determining roots
of equations.
13-2. SYNTHETIC DIVISION
A simplification of the ordinary method of long division, called
synthetic division, will be presented here. This abbreviated method
not only enables us to quickly find the quotient and remainder when
a polynomial f(x) = a Q x n + a^ x n ' 1 H h a n is divided by a binomial
of the form x r, but also affords a simple process for substituting
values of the variable into a polynomial.
We shall divide 2x 3 - 9x- + I3x + 5 by x - 3 to illustrate the pro-
cedure in synthetic division as compared with that of long division
considered in Section 1-19.
235
236 Theory of Equations Sec. 13-2
By long division we have :
2z 3 - Qx 2 + 13x
a- 3
2x 2 - 3x + 4
I3x
Qx
4z + 5
4a - 12
+ 17
Thus, the quotient is 2x 2 3x + 4, and the remainder is 17.
Since like powers of x are written in the same vertical column,
the work may be shortened by writing only the coefficients, as in
the following schematic arrangement:
1-3
2
2
-9
-6
+ 13
+ 13
+ 9
+ 5
+ 5
- 12
Q
-3
+ 4
+ 4
2-3+4
+ 17
Next, we note that the first term in the divisor x r need not be
written, since the divisor is always linear, and the coefficient of x
in it is always unity. Moreover, it is not necessary to write the first
term in each row that is to be subtracted, since its coefficient is
always the same as that of the term directly above. Also, only
the first term of each partial remainder needs to be written down,
for the second term is the same as the term directly above it in the
first row. Finally, the coefficients in the quotient need not be
written, since these are precisely the leading coefficient in the
dividend and the remaining partial remainders, excepting the last.
Hence, we may indicate the process in the following way :
2-9+13+5 | -3
-6
+ 9
- 12
+ 17
Sec. 13-2 Theory of Equations 237
This scheme can be written compactly as follows :
2 - 9 13 5 | - 3
- 6 9 12
2-3 4 17
If we replace -3 by +3 in the divisor and add the partial prod-
ucts in the second row instead of subtracting them, we obtain the
same result. The synthetic division then takes the following form :
2-9 13 5 | 3
6 -9 12
2-3 4 17
Here the numbers 2, -3, and 4 in the third row are the coefficients
of the quotient, and the last number 17 is the remainder.
We can now outline the procedure for synthetic division. Note
that in every step of the procedure, immediate reference is made
to the illustrative example.
To divide f(x) = a x n + a^ x n ~* H h a n by x r, first arrange
f(x) in descending powers of x, writing zero for the coefficient of
any missing power of x. Then arrange the numbers involved in the
process in three rows, as shown in the following steps :
Step 1. In the first row, write the coefficients in f(x) in order, as
a ( >, !, a n . At the right, put the constant term of the divisor with
its sign changed. We have then
Example
a ai a 2 a n | r 2 - 9 13 5 | 3
Step 2. Bring down the first coefficient a of f(x) into the first
place of the third row. Thus, we have
Example
i a2 a n \_r_ / 2 9 13 5 | 3
i;
Step 3. Multiply a by r, and write the product a r in the second,
row under ai. Bring down the sum of ai and a r into the third row.
Thus, we now have
Example
a ai a 2 a n |_r_ 2 - 9 13 5
apr 6 .
+ i) 23
238 Theory of Equations Sec. 13-2
In the example, multiply 2 by 3 and write the product 6 in the
second row below 9. Then add 9 and 6, writing the sum 3 in
the third row.
Step 4. Multiply o r 4- di by r, place the product in the second
row under a 2 , and add. Continue this process until finally a product
has been added to a n .
The complete solution of the illustrative example follows :
2-9 13 5 | 3
6-9 12
2-3 4 17
In the last operations, 3 is multiplied by 3, and the product 9
is written below 13. The sum of 13 and 9 equals 4. Finally the
product of 4 and 3, or 12, is written below 5 and added to a n 5 to
give 17.
Step 5. When the process is completed, the last number in the
third row directly below a n is the remainder. The other numbers
in this row, read from left to right, are the coefficients of powers of
x in the quotient arranged in descending order.
The entire synthetic process of dividing a polynomial f(x) by
x r, although it is somewhat complex notationally, can be con-
veniently exhibited as follows :
ao ai a2 a n _i a n [ r
bpr b\r b n ~2r b n -\r
bo hi 62 b n -i R
Here the expressions for the coefficients 6 , &i> b 2 , , b n _i of the
powers of x in the quotient are 6 = a , &i a () r + Oi, 6 2 = Oo r2
+ ai r + 02 , b n -i = do r 11 - 1 + a! r- 2 + + a n -i. Hence, the quotient
may be written
(13-1) q (x) = b x*- 1 + Bi x n ~ 2 + + & n -i-
Also, the remainder assumes the form
(13-2) R = a r n + air"- 1 H h a n ^r + a n .
The expression for R is precisely the result of substituting r for x
inf(x). In other words
(13-3) B=/(r).
Finally, if we subtract the product of x r and q(x) from f(x),
we obtain the remainder R = f(r). Or, if we transpose the product,
we have the usual statement found in discussions of division. That is,
(13-4) /(*) = (*-r).j(*
Sec. 13-2 Theory of Equations 239
The following examples will further illustrate the process of
synthetic division.
Example 13-1. Divide 3x* - 4z 2 + x - 2 by x + 2.
Solution: Since x r = x + 2, we have r = 2. Writing zero for the coefficient
of the missing power x 3 , we have the following result:
30-4 1-2 |-2
- 6 12-16 30
3-6 8-15 28
Note that the first -coefficient 3 is brought down into the first place of the third
row. Next 3 is multiplied by 2, and the product, which is 6, is written in
the second row under 0. The sum of and - 6, or - 6, is written in the third row
directly below 0. Proceeding in this way, we find that the quotient is 3x 3 fix 2 + 8x 15
and the remainder is 28.
Example 13-2. Given /(x) = z 3 - 2x 2 + 5x - 4, find
Solution: By (13-3), /(r) equals the remainder obtained in the division of f(x)
by x r. Hence, we have the following results :
a)
Therefore, /(- 1) = - 12.
Therefore, /(I) = 0.
c)
Therefore, /(3) = 20.
1
- 2
5
- 4
LJ
- 1
3
- 8
1
-3
8
- 12
1
-2
5
- 4
u
1
- 1
4
1
- 1
4
1
-2
5
- 4
UL
3
3
24
1
1
8
20
EXERCISE 13-1
In each of the problems from 1 to 15, divide the first function by the second, to
find the quotient and the remainder, by using synthetic division.
1. x 2 - Sx + 7, x - 1. 2. x* - x 2 - Sx + 6, x - 3.
3. x 2 - 5x + 6, x - 4. , 4. z 3 - Sx 2 + fix - 6, x - 3.
5. z 4 - 3x 2 - 6x + 6, i + 3. 6. x 3 - 3x 2 + Or - 24, x - 4.
7. 2x 4 - x 3 - 6z 2 + 4z - 8, x - 2. 8. 2x 3 + Sx 2 + 4, x + 2.
9. .c 3 + 3x 2 - 2,r - 5, x - 2. 10. 2x 5 - 3z 3 + 2z 4- 1, z + 2.
11. z 4 4- z 3 - 59s 2 - 69z + 030, z 2 - a? - 42. (Hint: Factor, the divisor and
divide successively by each factor.)
240 Theory of Equations Sec. 13-2
12. x* - 3x 3 + So; 2 - 3x + 2, x 2 - 3x + 2.
13. z 5 + 2x 4 - 21x + 18, a; 2 + 2x - 3.
14. a w 1, o 1.
15. x* y* y x y.
For each of the following polynomial functions, find the indicated values by the
method of synthetic division :
16. f(x) = 3x 3 - 7x2 _ 5x + 6 . Find /(I) and/(- 4).
17. /(x) = a;* - 2x 3 + 2x 2 - 5x + 2. Find/( - 2) and/(0).
18. /(a?) = x 4 - 4x 3 - 4x 2 + 24x - 9. Find/(3) and/(9).
19. /(x) = x* - 3x3 _ 13^2 + 21 X + 18. Find /(I) and/(3).
20. /(x) = 7x 4 + 37x3 _ X 2 _ 14a . + 4. Find /(I) and/(- 5).
13-3. THE REMAINDER THEOREM
In Section 13-2, it was shown that the remainder in the division
of a polynomial by a binomial x r can be found without actually
performing the division. Thus, in establishing (13-3), we have
proved the following theorem.
Remainder Theorem. If a polynomial f(x) is divided by x r,
the remainder is the value of f(x) for x = r; that is, the remainder
It follows from this theorem that f(x) is exactly divisible by
x r if and only if /(r) = 0. Hence, we have proved also the fol-
lowing theorem.
Factor Theorem. If f(r) = 0, then x r is a factor of the poly-
nomial f(x), and conversely.
Example 13-3. Is x + 3 a factor of x 4 - 2x 3 + 3x 2 - 5?
Solution: Here/(x) = x 4 - 2x 3 + 3x 2 - 5. Also, x - r = x + 3, and r = - 3.
According to the factor theorem, /( 3) must equal zero if x -f 3 is to be a factor
of /(x). But /(-3) = (-3) 4 - 2 (-3) 3 +3 (-3) 2 -5 =81 +54 +27 -5
= 157. Therefore, since /( - 3) 7* 0, x + 3 cannot be a factor of x 4 - 2x 3 + 3x 2 - 5.
Example 13-4. Given /(x) = x 3 - 3x 2 + 5x - 6. Show that /(2) = and,
therefore, that x 2 is a factor of /(x).
Solution: Since /(x) = x 3 - 3x 2 + 5x - 6, we have by substitution
/(2) = (2) 3 - 3 (2) 2 + 5 (2) - 6 = 0.
From the fact that/(x) equals zero when x = 2, it follows from the factor theorem
that x - 2 is a factor of /(x). The student should check this result by synthetic
division and find that
/(x) = x 3 - 3x 2 + 5x - 6 = (x - 2) (x 2 - x + 3).
Sec. 13-4 Theory of Equations 241
Example 13-5. Find under what condition (x -f a) is a factor of x* + a n , where
n is an integer and a ^ 0.
Solution: In this case, f(x) = r + a n , and/( - a) = ( - a) w + a n . This sum can
equal zero only if ( a) n = a n , that is, only when n is odd.
EXERCISE 13-2
In each of the problems from 1 to IS, determine if the second function is a factor
of the first. If it is a factor, find another factor.
1. 2z 3 - Gx 2 + x + 6, x - 2. 2. z 3 + 4z 2 -f bx + 6. x 2 + x + 2.
3. z 3 + 2z2 - 3z - 1, x - 1. 4. z 7 - 1, z - 1.
5. x 3 - z 2 - llx + 15, x - 3. 6. z 5 + 243, .r + 3.
7. x s - 256?/ 16 , z - 2?y 2 . 8. 2z 3 -f 3z 2 - 9z - 111, z - 1.
9. x 4 - 4z 3 - x 2 + 16.c - 12, z + 1.
10. .r 5 + 5s 4 + 20z 3 + GO* 2 + 120z + 120, x + 1.
11. 5z 3 + aft (5 - 6r?& 2 )z 2 - 2a 2 6 2 (5 + 3a6 2 )a: + 12a 4 fc 5 , x - ab.
12. 2x* - 3x* - 3z - 2, x + 2.
13. i - 10.r 4 + IS* 8 - 24.1* + 75, x - 2.
14. 2x 4 - 31x 3 + 21x 2 - I7x + 10, x + 1.
15. 12.x 4 - 4(Xr 3 - x 2 + lllj - 90, 2x - 3.
16. 12z 3 - 22.r 2 - 34z + GO, 3x + 5.
17. 24* 4 - 122x 3 + 159x 2 - Ix - GO, 3z - 4.
18. 24x^ _ 74^.4 __ 85.r 3 + 311x 2 - 74x - 120, 2x + 1.
19. Show that the equation x 4 + 2.r 3 - 9r 2 - 2x + 8 = has the roots 1,2, - 1, - 4.
20. Find all roots of the equation 12.r 4 - 4Qx* - x 2 + lllx - 90 = 0, given that
3/2 is a double root, that is, that (x 3/2) 2 is a factor of the left side.
21. Prove that a b is a divisor of a n b n for every positive integral value of n.
22. Prove that a + b is a divisor of a n - b n if n irf a positive even integer.
23. Prove that a + 6 is a divisor of a n -f- b n if n is a positive odd integer.
24. Prove that neither a + b nor a - b is a divisor of a n + 6 n if n is a positive
even integer.
13-4. THE FUNDAMENTAL THEOREM OF ALGEBRA
We usually assume that every algebraic equation with real or
complex coefficients has at least one real or complex root. Although
the existence of such a root is not to be taken for granted, a proof
is beyond the scope of this book. Accordingly, we shall accept the
following theorem as true.
Fundamental Theorem of Algebra. Let f(x) be a polynomial of
degree n with complex coefficients. Then the algebraic equation
f(x) =0 has at least one complex root.
The first complete and rigorous proof of the fundamental theorem
was given by Gauss in the beginning of the nineteenth century.
Since that time, many proofs have appeared, but most require
knowledge of the theory of complex functions.
242 Theory of Equations Sec. 13-4
By a repeated application of the fundamental theorem, it can be
shown that the number of roots of any polynomial equation with
real or complex coefficients is equal to the degree of the poly-
nomial. We shall state the following theorem and give its proof.
Theorem. If f(x) is a polynomial of degree n, the equation
f(x) =0 has exactly n roots.
Proof. Let f(x) be a polynomial of degree n with real or com-
plex coefficients. By the fundamental theorem, there is a number TI
such that f(ri) =0. Then, by the factor theorem,
f(x) = (x - n) qi(x),
where q\(x) is a polynomial of degree n 1 whose leading coeffi-
cient is OQ.
Likewise, q\(x) = has a root by the fundamental theorem. If
we denote this root by r 2 , then QI (r 2 ) = 0. Also,
qi(x) = (x - r 2 ) 92(3),
where tf 2 (#) is of degree n 2 with leading coefficient a .
Similarly, # 2 (#) = also has a root. We can continue in this way
until we come to a polynomial of the first degree with root r n . Then
(13-5) f(x) = a (z - n) (a - r 2 ) (a - r n ),
where a is the coefficient of x in our final first-degree polynomial.
Since f(x) equals zero when we substitute for x any one of the n
numbers r lf r 2 , , r, it follows that the equation f(x) = has at
least the roots r a , r 2 , , r n .
Moreover, there are no other roots. For, suppose that r is some
root other than n, , r n . Substitution of r for x in (13-5) yields
/(r) = a (r - ri) (r - r 2 ) (r - r n ).
The right side of this equation cannot equal zero, since no one of the
factors can equal zero. Therefore, /(r) ^0, and f(x) = has no
more than the n roots found before.
It may happen that a certain root appears more than once among
the numbers r t , r 2 , , r. In that case it will be counted as many
times as the corresponding equal factors appear in (13-5). If a
certain factor (x - r) appears m times in (13-5), then r is said to
be a root of multiplicity m. A root is called a simple root, a double
root, a triple root, and so on, the proper name depending on how
many times the same factor appears. Hence, combining the state-
ments that "f(x) = has at least n roots' 1 and "f(x) = has no
more than n roots," we can conclude that a polynomial equation of
the nth degree has exactly n roots, a root of multiplicity m being
counted as m roots.
Sec. 13-5 Theory of Equations 243
It is to be noted that a rigorous proof of this theorem requires the
use of induction. (See Chapter 16.)
13-5. PAIRS OF COMPLEX ROOTS OF AN EQUATION
We wish to remind the student that while an equation of the nth
degree has n complex roots, the number of real roots may be less
than the degree of the equation. For example, x 2 + 1 = has no
real roots. Determination of the number of real roots may be sim-
plified by use of the following theorem.
Theorem. If all the coefficients of f(x) = are real numbers, and
if the complex number a + bi is a root of f(x) = 0, then the conju-
gate a bi is also a root. It is understood that a and b are real and
6^0.
Proof. Let x in f(x) = a x n + a x x n ~ l H h a n be replaced by
a + bi. Then we have
f(a + bi} = a (a + bi) n + ai(a + bi)*- 1 + + .
If we expand the powers of a + bi by the binomial theorem and
simplify the resulting expression, then all terms which contain even
powers of i will be real, while all terms which contain odd powers
of i will be pure imaginary. Denote by P the aggregate real part,
and by Q the aggregate imaginary part. Then we have
/(a + bi) = P + Qi = 0.
Hence, in accordance with (11-4) , P = Q = 0.
Now, replace x by a bi in f(x) 0. In those terms of f(a bi)
in which bi is raised to an even power, the result will remain the
same as in f(a + bi) . However, all terms in f(a bi) in which bi
is raised to an odd power will have their signs changed. Hence,
/(a - bi) = P - Qi.
But, since we have shown that P = Q = 0, we conclude that
P-Qi = 0.
In other words, a bi is also a root of f(x) = 0.
Example 13-6. Solve x* - x 3 - 2x 2 + 6s - 4 = 0, one root being 1 + i. ;
Solution: According to the last theorem, both 1 + i and 1 i are roots of the
given equation. Using Eqs. (12-3) and (12-4) for the sum and product of two
roots, we find that these conjugate complex numbers are roots of a: 2 2x +2 = 0.
Dividing the original polynomial by this quadratic function, we get the quotient
x 2 + x - 2. Solution of the equation x 2 + x - 2 = yields the remaining desired
roots, x = - 2 and x = I.
244
Theory of Equations
EXERCISE 13-3
Sec. 13-5
1. Solve X s + x 2 .- 2x -f 12 = 0, one root being 1 + V3f.
2. Solve x 3 + 3z 2 + I2x - 16, one root being -2 - 2 V~ 3.
3. Solve z 4 - 2x 3 - 7x 2 + ISx - 18 = 0, one root being 1 - f.
4. Solve x* -f 3z 3 -f 7z 2 -f 6z + 4 = 0, one root being - 1 - Vi.
5. Solve a; 5 - 8z 4 -f 27z 3 - 46z 2 + 38x - 12 = 0, one root being 2 - vX
and one root being 1.
6. Solve x* + x 2 + 1 = by considering the equation to be a quadratic equation in x 2 .
7. Find a real cubic equation, two of whose roots are 2 and 1 -f 2i.
8. Find a real equation of lowest degree having the roots i and 1 - i.
13-6. THE GRAPH OF A POLYNOMIAL FOR LARGE VALUES OF x.
In graphing a polynomial function, it is helpful to know the
location of points on the curve for numerically large values of x. It
can be shown that, when x is numerically sufficiently large, the term
a x n of highest degree is numerically larger than the sum of all the
other terms combined. Therefore, the sign of this term determines
the sign of the entire polynomial.
Let us consider the values of the function f(x) = a; 3 + 5x 2 - Ix
13 as x assumes various values from left to right along the #-axis,
that is, for increasing values of x. The results may be tabulated
conveniently as follows :
X
x 3
5x 2 - Ix - 13
x s + 5x 2 Ix
13
- 10
- 1000
500 + 70
- 13 =
557
- 443
- 8
- 512
320 + 56
- 13 =
376
- 136
- 6
- 216
180 + 42
- 13 =
209
- 7
0+
- 13 = -
13
- 13
5
125
125 - 35
- 13 =
77
202
10
1000
500 - 70
- 13 =
417
1417
When x is negative, but numerically sufficiently large, it is seen
that f(x) is negative. Thus, when x = - 10, x* = - 1000, while the
sum of other terms, or the value of 5x 2 -Ix - 13, is only 557 ; and
/(-10) = -443. For points far enough to the right of the origin,
say for x = 10, f(x) is positive. For example, /(10) = 1000 + 417 =
1417. Hence, the graph of y = f(x) would be located below the
#-axis on the left but would rise above the #-axis as x gets larger
and larger.
As a second illustration, let us consider the function f(x) =
x 4 + 7x 2 -8x + 10. Here f(x) is positive for large numerical values
of x, regardless of the sign of x. Hence, in this case the graph
Sec. 1 3-8 Theory of Equations 245
would be above the re-axis for large numerical values of x to both
right and left of the origin.
The following helpful conclusion can be drawn from the preced-
ing discussion involving large values of x.
When x is sufficiently large and positive, f(x) has the same sign
as the leading coefficient a. When x is negative and sufficiently
large numerically, f(x) has the same sign as a when n is even, and
has the opposite sign when n is odd.
The following symbols are sometimes found in discussions of the
values of functions for numerically large values of x. When the
symbolic statement /(-foo)>0is used, what is meant is that, for
all sufficiently large positive values of x, f(x) ig positive. Similarly,
the statement /(+ oo) < means that for all sufficiently large posi-
tive values of x, f(x) is negative; the statement /( oo) > means
that, for all negative values of x which are sufficiently large numer-
ically, / (x) is positive ; and / ( co ) < means that, for all negative
values of x which are sufficiently large numerically, f(x) is negative.
Thus, if f(x) =x* + 5x 2 -7x- 13,/(- oo) <0 while /(+ oo) > 0.
However, if f(x) = x* + 7x 2 - Sx + 10, /(- oo)>Oand/(+ co)>0.
13-7. ROOTS BETWEEN a AND b IF f(o) AND f(J>) HAVE OPPOSITE SIGNS
Another helpful theorem relating to the roots of a polynomial
equation is the following.
Theorem. If the coefficients of a polynomial f(x) are real, and if
a and b are real numbers such that f(a) and f(b) have opposite
signs, then the equation f(x) = has at least one real root between
a and b.
We shall not give a proof of this statement here, but shall merely
mention the following geometric considerations. The graph of a
polynomial is a continuous curve; that is, it has no "breaks." There-
fore, if the points (a, /(a)) and (b,f(b)) lie on opposite sides of
the #-axis, the graph apparently has to cross the #-axis at least
once between these points.
13-8. RATIONAL ROOTS
The following theorem is fundamental for the solution of equa-
tions having integral coefficients.
Theorem. If the equation
f(x) - a x n + aix n ~ l H h a n =
with integral coefficients has the rational root -, > where c and d
a
246 Theory of Equations Sec. 13-8
are integers having no common factor > 1, then c is a divisor of
the constant term a n , and d is a divisor of the leading coefficient a .
Proof. We shall make use of a principle from the theory of num-
bers: If an integer c divides the product of two integers a and 6
and if 6 and c have no common divisor other than 1, then c is
a divisor of a.
XI
Let -3 be a root of f(x) = 0, where c and d are integers with no
a
common divisor other than 1. Then
c n c n ^ c
a ^ + ai d^r + ' " + a "~ l d + a " = a
Multiplication by d n gives
aoc w + aic n ~ l d H h a n -icd n ~ l + a n d n = 0.
Since c divides all terms before the final one, c also divides that
term. If now c is factored into primes, none of these primes is a
divisor of d, and therefore of d n . Thus each prime divides a n , and so
c itself divides a n . In a similar fashion, it may be shown that d
divides a .
Example 13-7. Find the rational roots of 3z 3 - I7x 2 + I5x + 7 = 0.
st
Solution: The possible rational roots are of the form -r > where c is a divisor of
ct
the constant 7 and d is a divisor of the coefficient 3. The only possible roots are
1, 7, 1/3, 7/3.
Using synthetic division to check 1, we obtain the following result:
3 - 17 15 7 |- 1
- 3 20 -35
3 -20 35 -28
By the remainder theorem, /( - 1) = - 28. Hence, - 1 is not a root.
We see, however, that /(O) = 4-7. Therefore, by the theorem in Section 13-7,
there must be a root between x = 1 and x = 0. Examining our list of possible
roots, in the hope that a rational root lies between 1 and 0, we see that a pos-
sibility is 1/3. The check by synthetic division follows:
3 - 17 15 7 |- 1/3
-16-7
3 - 18 21
Hence, x = 1/3 is a root. After dividing the given polynomial by x + 1/3, and
equating the quotient 3x 2 18z 4- 21 to 0, we obtain the quadratic x 2 6x + 7 = 0.
This equation has the roots 3 =h \/2, which are not rational numbers. Therefore,
the only rational root is = 1/3.
In this example, as is sometimes the case, it has been possible to find all the roots
of the given equation.
Sec. 13-8 Theory of Equations 247
EXERCISE 13-4
1. Show that l&r 3 - 33z 2 + 2x + 5 = has real roots between - 1 and 0,
between and 1, and between 1 and 2. Find these three roots.
2. Find the rational roots of 2x 3 - 9z 2 + 3x + 4 = 0.
3. Prove that x 3 + 2x 2 3x - 5 = has at least one positive root.
4. Prove that x* x 3 4- x 2 + x 3 = has at least one positive root and at
least one negative root.
5. Prove the following corollary of the theorem in Section 13-8. If f(x) =
x n + aix*- 1 + + a n = has integral coefficients and has an integral
root r, then r is a divisor of a n .
6. Find the integral roots of x* - 1 = 0.
7. Show that the equation x 2 + x + 1 = has no rational roots.
8. Solve the equation x 3 1 = 0.
9. Find all the integral roots of a: 8 + z a + x + I = 0.
5 3
10. Show that 3 is a root of x 3 - ^ x 2 - 2x + - = 0. Why does this not contra-
z &
diet the theorem in Section 13-8 or that in Problem 5?
In each of the problems from 11 to 20, find all roots of the given equation.
11. x 4 - Sx 2 + 16 = 0. 12. 4x 3 - 16z 2 - 9z + 36 = 0.
13. 3x* + x 2 + x - 2 = 0. 14. 2x* + 3z 2 - 6x - 9 = 0.
15. 2z 3 - x 2 + 2x - 1 = 0. 16. x s - llz 2 + 37x - 35 = 0.
17. 5z 3 - 13z 2 + 16s - 6 = 0. 18. x* - Sx 3 + 37z 2 - 50z = 0.
19. 2x* - x 2 - 4z + 2 = 0. 20. 3z 3 - 13z 2 + 13z - 3 = 0.
14
Inequalities
14-1. INTRODUCTION
In previous chapters we have explained some methods of deter-
mining the roots of an equation. By applying these methods, one
can find the values of an unknown for which a certain function of
the unknown equals zero. Often, however, it is necessary to solve an
inequality, that is, to discover for what values of the unknown a
certain function is less than or greater than another function.
The present chapter is concerned primarily with the solution of
inequalities, and the following discussion is essentially an extension
of the study of the order relation undertaken in Section 1-8. We,
therefore, recommend that the student thoroughly review Section
1-8 before starting the study of the present chapter. Since the solu-
tion of inequalities often involves the use of absolute values, a thor-
ough mastery of Sections 1-9 and 1-10 is also a requirement.
The classification of inequalities corresponds to that of equalities
or equations. As in the study of equations, there are two kinds of
inequalities involving unknowns, namely, absolute inequalities and
conditional inequalities.
An absolute inequality is an inequality that is satisfied by all
values of the variable or variables for which the functions appear-
ing are defined.
A conditional inequality is one that is true only for certain values
of the variable or variables. Thus, x 2 + 1 > (where x is real) is
an absolute inequality, because it is true for every real value of x ;
but x 1 > is a conditional inequality, because it is valid only
when x > 1.
14-2. PROPERTIES OF INEQUALITIES
The rules for dealing with inequalities are to some extent analo-
gous to those for equations. In transforming inequalities, we shall
have occasion to use the following elementary principles which
248
Sec. 14-3 Inequalities 249
follow at once from the fundamental properties proved in Section
1-8.
Principle 1. If a < &, then a c < b c.
Here are three illustrations :
From 12 > 8, it follows that 12 + 3 > 8 + 3.
From 8 < 12, it follows that 8 - 2 < 12 - 2.
From 12 > 8, it follows that 12 - 8 > 0.
Principle 2. If a < b and c> 0, then ac < be and - < -
c c
Two illustrations are given here :
From 3 < 5, it follows that 3 2 < 5 2.
Q 1 A
From 8 < 10, it follows that < ~
,
Principle 3. If a < b and c < 0, then ac > be and - > -
c c
Here is an illustration :
From 3 < 4, it follows that -3 > -4.
14-3. SOLUTION OF CONDITIONAL INEQUALITIES
The process of solution of a conditional inequality consists in
finding all values of the variable which satisfy the inequality. A
solution consists of a set of values of the variable, rather than one or
more isolated values as is usual in the case of a conditional equation.
The discussion in this section is limited to inequalities involving
rational functions in only one variable. If this variable is x, the
inequality can be written in the form f(x) > or f(x) < 0. For
instance, suppose we want to solve the inequality
x 2 - x > 2.
We may then obtain the following equivalent inequality:
/(a) = x 2 - x - 2 > 0.
It is easily seen that this transformed inequality has the same solu-
tion set as the original inequality.
In solving this transformed inequality, we find first the values
of x, if there are any, for which f(x) changes sign as x increases
in magnitude. If f(x) is a polynomial, such a change of sign occurs
when f(x) =0. In the example under consideration, changes of
sign are obtained only at points where
/GO = x 2 -x-2=Q.
Hence, we must find the roots of x 2 - x - 2 = 0. These are 1 and
2, and they determine on the #-axis three intervals throughout each
of which f(x) retains the same sign. In other words, to find the
250 Inequalities Sec. 14-3
solution of f(x) > 0, we find the interval or intervals within which
f(x) has the sign indicated in the given inequality. In the example
under consideration this sign is positive, because f(x) is to be > 0.
The method is applied in Example 14-1.
In general, the solution of an inequality is obtained by equating
the function f(x) to zero and solving the resulting equation. If the
inequality is of the form i?i ^ 0, where p(x) and q(x) are poly-
ps)
nomials, it may be cleared of fractions by multiplication by [q (x) ] 2 .
Since the square of any non-zero real number is positive, the sense
of the inequality is not changed by multiplication by this factor.
This leads to the form f(x) ^ 0, where f(x) is a polynomial. The
values of x for which f(x) changes sign are called critical values.
When the critical values are arranged in increasing order, they
determine on the #-axis intervals, throughout each of which f(x)
cannot change sign. Consequently, the required solution is repre-
sented by the set of values of x for which f(x) has the same sign
as that indicated in the given inequality.
Note. In general, it can be shown that if a factor of f(x) appears
to an odd power (that is, if f(x) = has roots of odd multiplicity),
the function will change sign at values of x for which this factor
vanishes. If a factor of f(x) appears to an even power (that is, if
f(x) =0 has roots of even multiplicity), the function will not
change sign at values of x for which this factor vanishes. There-
fore, it is sufficient to test only one value of x in one interval. This
test gives the sign off(x) in that interval. The sign of f(x) in each
of the other intervals can be quickly and easily determined from
the multiplicity of the critical values. Substituting into f(x) a value
of x in each interval then provides a check of the solution.
Example 14-1. Solve the inequality
x 2 x > 2.
Solution: An equivalent inequality is
f(x) = x 2 - x - 2 > 0.
From the preceding discussion it follows that the critical values are the roots of
x 2 - x - 2 = 0.
These roots are - 1 and 2.
As shown in Fig. 14-1, the points 1 and 2 determine on the z-axis the following
three intervals
(a) x < - 1; (b) - 1 < x < 2; (c) x > 2.
-2
-1 1
2 3
j ^
^/
W
W
FIG. 14-1.
Sec. 14-3
Inequalities
251
Throughout each of these intervals, f(x) retains the same sign. This condition
may also be seen from the graph of y = x 2 x 2 shown in Fig. 14-2. Here
we note that in each of these intervals the curve lies either entirely above the
a>axis or entirely below it.
The solution of f(x) > can now be found by examining the sign of /(a?) in each
of these intervals. Thus, for a value such as x = - 2 in the interval (a), /( - 2)
= (- 2) 2 - (- 2) -2=4. Hence, f(x) is positive throughout the interval (a).
In the graph of y = x 2 x 2 shown in Fig. 14-2, the curve lies above the
x-axis to the left of x = 1.
For the value x = in the interval (6),
we have /(O) = - 2. Hence, f(x) is nega-
tive throughout this interval, and the graph
lies below the g-axis.
For the value x = 3 in the interval (c),
/(3) = (3) 2 - (3) - 2 = 4. So/(.r) is posi-
tive, and the graph again lies above the
.r-axis.
These same results may also be obtained
by the following much shorter procedure:
Select a value of x in the interval (6) for
which /(x) is easily evaluated. Such a value
isz=O.Since/(0) = -2,/(x) <0 throughout
this interval. Therefore, f(x) > for inter-
vals (a) and (c), because the sign of f(x)
changes at the critical values x = 1 and _ * A
x=2.
We see, therefore, that in the intervals (a) and (c), f(x) has the sign indicated by
the given inequality. Since f(x) must be greater than zero, the solution set of the
given inequality is described by x < - 1 and x > 2.
Example 14-2. Determine the values of x for which \/x 3 2x 2 - 3x is real.
Solution: We shall solve the equivalent problem
f(x) = z 3 - 2z 2 - 3x ^ 0.
Solving the equation x 3 2x 2 3# = 0, we find that the critical values are 1, 0, 3.
(d)
As shown in Fig. 14-3, these critical values determine on the #-axis the following
four intervals:
(a) x < - 1; (6) - 1 <x < 0;
(c) < x < 3; (d) x > 3.
Throughout each interval f(x) has the same sign.
In this example we may select z = 1 in the interval (c). Since /(I) = (I) 3 2(1) 2
- 3(1) = - 4, f(x) < throughout this interval.
252 Inequalities Sec. 14-3
As we proceed into the interval (6), we find that/(z) changes sign when x = 0.
Therefore, f(x) > in the interval (6). Again f(x) changes sign when x = - 1
and becomes < in the interval (a). Proceeding to the right from the interval (c)
into the interval (d), we find that/(#) changes sign when x = 3 and is > in the
interval (d).
Thus, we have the following results:
in (a), /(a) < 0; in (&),/(*) > 0;
in (),/(*) <0; in(d),/(x) > 0.
Hence, the inequality /(x) > is satisfied in intervals (b) and (d). And, since the
condition x 3 2x 2 3x = is also allowed in the original problem, the values
1, 0, 3 are included in the solution. Therefore, the solutions for f(x)
= x 3 2x 2 3x ^ are 1 ^ x ^ and x ^ 3. These are the values of x for
which the original expression V# 3 2# 2 3x is real.
x 3 3x 2
Example 14-3. What values of x satisfy ^- > 0?
X t
Solution: Clear the given inequality of fractions by multiplying by (x 2) 2 and
obtain f(x) = (x - 2) (& 8 - 3rc 2 ) > 0. Solving the equation z 2 (,r - 2) (x - 3) = 0,
we find that the critical values are 0, 0, 2, 3.
Hence, we have the, following four intervals, as shown in Fig. 14-4:
(a) x < 0; (b) < x < 2; (c) 2 < x < 3; (d) Z > 3.
Throughout each of these intervals /(x) has the same sign.
Let us initially test f(x) for z = 1 in the interval (b). Since /(I) = (1 - 2) [(I) 3
- 3(1) 2 ] = ( - 1) ( - 2) = 2, it follows that/Or) > throughout this interval.
We see that /Or) does not change sign for the critical value x = 0, because x
is a double root of f(x) = 0. Hence, f(x) > in the interval (a).
As we proceed to the right from the interval (b) the f unction f(x) changes sign for
each of the critical values x = 2 and x = 3. Hence, f(x) < in the interval (c),
and f(x) > in the interval (d).
Therefore, f(x) is positive in the intervals (a), (6), and (c?). That is, the solution
set of the original inequality is described by x < 2, excluding the value x = 0,
and x > 3.
Example 14-4. Solve the inequality x 2 2x + 3 > 0.
Solution: Let f(x) = x 2 2x + 3. Since the roots of f(x) = are imaginary,
there are no critical values. Hence, the graph of y = f(x) lies either entirely above
the x-axis or entirely below that axis.
Testing for x = 0, we find that/(0) = 3. This result indicates that the graph lies
above the z-axis. Consequently, the inequality is" satisfied for all real values of x.
Sec. 14-3 Inequalities 253
2x - 1 1
Example 14-5. Solve the inequality
Solution: By Section 1-10, the inequality | x - 6 | < a is equivalent to
3~
b-a<x<b+a. Hence, r I < 1, which is equivalent to | 2z 1 1 < 3,
may be written as follows :
l-3<2x<l+3 or - 2 < 2x < 4.
We thus find that the solution set of the original inequality is described by
- 1 < x < 2.
Alternate Solution: We may proceed by solving individually the two inequalities
2x 1 2x 1
1 < 5 and 5 < 1 and determining the common solutions; For
O O
2x 1 2x 1
1 < - i we have 3<2x 1, or 1 < x. For r < 1, we have
O o
2x - 1 < 3, or x < 2. The common solutions satisfy the inequalities - 1 < x and
x < 2. So we again have - 1 < x < 2.
EXERCISE 14-1
In each of the following problems, solve the given conditional inequality or
inequalities.
1. x - 3 < 0. 2. x + 1 < 0. 3, x + 5 > 0.
4. x - 1 0. 5. 4x - 16 < 0. 6. 4x - 16 > 0.
7. 6r + 3 < 0. 8. 4x - 8 > 3x - 10. 9. 4x < - 3x - 7.
10. 3 - 4x < 2x + 1. 11. - 3 < 6x < 3. 12. | x - 1 | < 3.
13. < 2-i + i < 1. 14. - 1 < ^? + ? < 1. IB. | 2* - 3 | < 4.
18. x 2 > 144.
19. 2x 2 32. 20. ^ - I < 0. 21. x(3x +2) < 1.
22. 4z 2 + 5x < - 1. 23. ^ < 3 _ 1 . - 24. ^ < - - - 1.
X 2 X X 2 X
25. 3ic 2 3# < 4. 26. 3# 2 4- 6a? < 9.
27. (x + !)(*+ 2) (x + 3) < 0. 28. (x - 4) (x + 5) (x - 6) > 0.
29. (x - 1) (x - 2) (x - 3) > 0. 30. (2x - 1) (x + 2) (3x 4- 1) ^ 0.
\J *
"** O Q
i. i**. -
x ^ 2
X
2
3
5
<2.
17. x
2 < 169.
31. Vz 2 - 25 is real. 32. \/x* - 5z + 6 is real.
33. > o. 34.
x + 2
35.
x +1
254 Inequalities Sec. 14-4
ABSOLUTE INEQUALITIES
To prove the truth of an absolute inequality, one must use the
known properties of the order relation. When none of these seems
readily applicable to the given inequality, it may be helpful to
replace this inequality by an equivalent one which may be more
easily treated. Repeated replacements may have to be made.
In carrying out a sequence of replacements, one need not verify
equivalence at each stage, provided that the final inequality can be
shown to imply the original one.
The methods for proving absolute inequalities may be used also
to prove theorems involving inequalities, as in Example 14-7.
Example 14-6. Prove that a 2 + 6 2 ^ 2ab for all real numbers a and b.
Solution: The given inequality is equivalent, by Principle 1, Section 14-2, to
a 2 - 2ab + b 2 0,
that is, to
(a - 6) 2 0.
This last inequality is true, because the square of every real number is non-
negative. Therefore, the original inequality is true also.
(I C
Example 14-7. If a, 6, c, and d are distinct positive real numbers, and if T < -5 >
. , .a a + c c
prove that -r < r . , < 3
o b + d, a
Solution: The inequality T < , , is equivalent, by Principle 2, Section 14-2, to
ab -h ad < ab + be.
This inequality is equivalent, by Principle 1, to
ad < be.
This inequality is true because it is equivalent to the given condition T < -5 by
Principle 2. Therefore, the inequality T < , , is true also.
Similarly, the inequality , , < -3 is equivalent, by Principle 2, to
ad -f ed < be -f ed,
and this inequality, by Principle 1, is equivalent to
ad < be.
a c
This last inequality again is equivalent to the given condition T < 3 Hence,
CL I / /*
the inequality , < ^ is true also. Therefore, it follows that the original
,... a a -f c c
inequalities T < , . , < j are true,
o o + a a
Sec. 14-4 /nequcrfiffes 255
EXERCISE 14-2
1. Prove that a 2 +1 2t 2a if a is real. Note that a 2 + 1 = 2a only when a = 1.
2. Prove that ^^ ^ ^TT if a > and 6 > 0.
2 a + 6
3. Prove that a 2 > a if a > 1 ; and that a 2 < a if < a < 1.
4. Prove that a > a 3 if < a < 1.
5. Prove that T + - > 2 if a and b are positive and a 9* b.
o a
6. Prove that a 2 + b 2 + c 2 ^ a& + fcc + ca. (Hint: From Example 14-6,
a 2 + b 2 2a6, 6 2 + c 2 ^ 26c, and c 2 + a 2 ^ 2ca. Add these inequalities.)
7. If a and 6 are two positive real numbers, the quantities - > \^ab, and ~-r
are called, respectively, the arithmetic mean (A), the geometric mean (Cr), and the
harmonic wean (H) of a and b. Prove that H <G < A, except when a = 6.
(In this case, we have A = G = //.)
8. Prove that a 3 + 6 3 > 3ab(a 6), if a and 6 are positive and a > 6.
9. Prove that ab + cd ^ 1, if a, 6, c, and d are positive, and if a 2 + & 2 = 1 and
C 2 -I- rf2 = 1.
10. Prove that 7 ^ T according as a $ 6, if a, 6, and c are positive.
b + c o
11. Prove that a 3 6 + ab 3 < a 4 + 6 4 , if a ^ 6.
12. If 2 = x + yi is a complex number, the modulus or absolute value of z is
denoted by | z \ = \/x 2 4- y 2 . Show that \z l + z 2 \ \zi\ +\z*\ and that
I z\ %2 1 ^ | 21 | | 22 | , for all complex numbers z\ and 22.
13. If 21, z 2 , and z 3 are complex numbers, prove that | Zi +z 2 \ \z\ +23 1 -h |2 -2 3 |.
14. If z is a complex number, prove that | | 2 | +| <2||.
15
Progressions
15-1. SEQUENCES AND SERIES
Sequences. An infinite sequence, called more simply a sequence
or sometimes a progression, is a single-valued function whose
domain of definition is the set of positive integers. A finite sequence
is a single-valued function whose domain consists of the integers
1, 2, , m for some positive integer m.
In specifying a sequence, it is necessary to give a definite rule of
correspondence which assigns to each integer n a single definite
number, or term, of the sequence. This term may be denoted by a n .
In particular, there is a first term ai corresponding to the integer 1,
a second term a 2 corresponding to the integer 2, and so on. The
sequence may be specified by the array of numbers
(15-1) aij a,2, as, , a n , .
This sequence is denoted briefly by {a n }. (As usual, throughout
this discussion, a row of dots stands for numbers assumed to be
present but not written.) A finite sequence has a "last" term and
may be designated by a i9 o- 2 , a m or simply by {a n }.
The nth term, or general term, of a sequence is denoted by a n .
From the rule specifying the nth term for each n f we obtain the
first, second, third, and other terms of the sequence by substituting
for n the values 1, 2, 3, and so on in turn. For example, if a n = 1/n,
the sequence is
111
lig'a'i""'
We should note that {a n } is the symbol for the sequence or func-
tion as a whole, whereas a n is the symbol for the nth term or value
of the function corresponding to the integer n.
256
Sec. 15-1 Progressions 257
There are many methods for specifying the function in the defini-
tion of a sequence. Two of these methods follow :
Explicit Formula. In one method, the nth term is given in terms
of n itself by means of an explicit formula.
Here are a few illustrations.
If a n = n, the sequence is 1, 2, 3, .
T , n ., .123
If a n = 2 _j_ ' ti 16 se( l uenc e is ~ > = 9 -^ i .
Tl ~T~ L Zi O 1U
T* n ^ .,23
If a n = jr > the sequence is 1, - > ^ > .
^71 1 ' o O
Recursion Rule. In another method, one or more of the first sev-
eral values of a n are given explicitly, and a rule is then given
whereby a n can be calculated from some or all of its predecessors.
A few illustrations are given here.
Let a n+ i = a n 2 + I with a\ = 0. Then
a 2 = ai 2 + 1 = O 2 + 1 = 1,
as = a 2 2 + 1 = I 2 + 1 = 2,
a 4 = a 3 2 + 1 = 2 2 + 1 = 5,
Let a n +i = (n + l)a n with a\ = 1. Then
a 2 = 2ai = 2 1 = 2,
a s = 3a 2 = 3 2 = 6,
a 4 = 4a 3 = 4 6 = 24,
Note that in this example a n = nl
Let a n +2 = <Wi + o n with ai = and 02 = 1. Then
as = fife + ai = 1 + = 1,
a 4 = a 3 + 02 = 1 + 1 = 2,
05 = 4 + 3 = 2 + 1 = 3,
Series. Let {a n } be a given sequence of terms ai, 03, , a*, .
Form a new sequence {s n }, where s n is obtained by adding the first
n terms of {a n }. The sequence of partial sums is then given by
Si = Oi, 52 = Oi + 02, ' ' >
s n = ai + 02 + + a n , (n = 1, 2, 3, )
The sequence {s n } formed in this way is called the (infinite) series
based on the given sequence {a n }.
The series as just defined is usually written in the following
abbreviated form:
(15-2) ai + a 2 +
258 Progressions Sec. 15-1
Two illustrations of series follow.
Illustration 1. Let a n = - and s n = ai + a^ H I- a n . Then the
n
partial sums are given by
si = ai = 1,
1 , 1 3
2 = Oi + a 2 = 1 + o = 9 >
1 1 11
The series may then be written
Illustration 2. Let a n = 2 4. an( ^ $n = &i + #2 + + <&n- Then the
partial sums are given by
112
= ai + a 2 = g + g - 3
The series is
2 6 12 n 2 + n '
Limit of a Sequence. One of the most important questions relat-
ing to sequences is whether or not a given sequence { a n } has a limit
as n increases indefinitely. If such a limit exists, the sequence
is said to be convergent or to converge to the limit <. Symbolically
this statement may be expressed as follows :
lim a n = .
w-oo
This notation is read, "the limit of a n as n increases indefinitely
is ."
In order to help make the concept of limit clear, we shall consider
the sequence given by a n = - The terms are
71
1 1 1
dl = 1, d2 = H ' a 3 = 7j >'> 0n =-
Z O fl
When we examine these terms, we notice that the larger the number
n is, the smaller the term becomes. In other words, as n increases,
the closer to zero is the number - In fact, - can be made as small
n n
as we please, if we merely choose n sufficiently large. For example,
Sec. 15-1 Progressions 259
- < - for every n larger than 100. We see also that - <
n 100 n 1000
for every n larger than 1000. We may conclude that for an arbi-
trary real positive number d, we can find a value of n, say any
integer N ^ ^ > such that for all integers n > N, it is true that
a
- < d. The limit and the sequence <-> are therefore related as
n (nl
indicated by the following statement.
The sequence < - 1 converges to the limit < = 0, if to each arbi-
(n)
trary positive number d there corresponds an integer N > such
that 0-d<-<0 + dfor every n > N.
n
In general, the limit of a sequence may be defined as follows :
Definition of Limit of Sequence. A sequence {a n } converges to the
limit , if to each arbitrary real number d > 0, there corresponds
a positive integer N such that d < a n < + d for every n > N.
This definition may also be put as follows :
Definition. A sequence { a n \ converges to the limit <, if for each
number d > there exists a positive integer N such that
| n | < d for every n > N.
Convergence of a Series. We shall again consider the infinite
series
(15-2) ai + a 2 + - - + a n + ,
which is the sequence of partial sums
Slj 52, * ' ' , S n , ' ' '
of the sequence {a n }. By the following definition the series is con-
vergent if the sequence of partial sums is convergent.
Definition. If the sequence of partial sums of the infinite series
(15-2) converges to a limit, and if lim S n = Sj then the series is
n- CD
said to converge to the limit S, and S is called the sum of the infinite
series.
The new use of the word "sum" for the value S of an infinite
series is perhaps unfortunate, for it seems meaningless to talk about
adding up the terms of an infinite series. Actually, S is not a sum,
but it is rather the limit of a sequence of partial sums of the series.
If a series does not converge to a limit as n becomes infinite, we
say that it is divergent, or that it diverges.
260 Progressions Sec. 151
EXERCISE 15-1
1. Given a\ = 1 and a+i = n + a n . Find the five terms of the sequence {a n }.
2. Given a l =4, a 2 = 3, and a n + 2 = 2a n +i a n . Find the first six terms of
the sequence { a n } .
3. Given ai = 1, a 2 = 2, and a n+ 2 = (n + l)a n+ i na n . Find the first six terms
of the sequence { a n } .
4. Show that the sequence a n = 3 n satisfies the recursion rule a n+ 2 = fln+1 + 6a n .
5. Show that the sequence a = ( 2) w also satisfies the recursion relationship
of Problem 4.
6. Assuming that A and B are any real numbers whatsoever, show that the
sequence a n = A3 n + #( 2) n satisfies the relationship of Problem 4.
7. Given a n +i = 2a n + 1 with ai = 2. Let s n = i + 2 + + a n . Find
ai, 02, , 05 and s lf s 2) , s.
8. Show that if s ai + a% + + , then s n+ i s n = a n +i
9. Prove that cr n = s n s w -i for n > 1, given ai = SL
10. If s n = n 2 , show that a n = 2n 1 and therefore that 1 + 3 + 5 +
+ (2n - 1) = n 2 .
15-2. ARITHMETIC PROGRESSIONS
An arithmetic progression is a sequence of numbers in which
each term after the first is obtained from the preceding one by
adding to it a fixed number; this number is called the common
difference.
Note that the common difference may be found by subtracting
any term of the sequence from the one that follows. Thus, 1, 5/2,
4, 11/2, is an arithmetic progression with the common difference
3/2, since 11/2 -4 = 4-5/2^5/2-1 = 3/2. Also, 5, 1, -3, -7 is
an arithmetic progression with the common difference 4, since
-7 -(-3) =-3-l = l-5 = -4.
It follows, therefore, that a necessary and sufficient condition
that three numbers A, B, and C form an arithmetic progression is
C-B^B-A.
15-3. THE GENERAL TERM OF AN ARITHMETIC PROGRESSION
Let a denote the first term of an arithmetic progression, and let d
denote the common difference. Then, by definition, an arithmetic
progression with n terms may be written as follows :
a, a + d, a + 2d ; a + 3d, , a + (n - l)d
Hence if l n represents the value of the nth term,
(15-3) l n = a + (n - l)d.
Sec. 15-4 Progressions 261
Also, we may write an arithmetic progression of n terms in the
following manner :
a, a + d, a + 2d, - , l n - 2d, Z n - d, t.
We shall be concerned with five quantities in connection with an
arithmetic progression. These are the first term a, the number of
terms n, the nth term l n , the difference d, and the sum S n of the n
terms.
15-4. SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION
Let S n represent the sum of the first n terms of an arithmetic
progression. If we write the indicated sum in both direct and
reverse orders, we have
S n = a + (a + d) + (a + 2d) + + (l n - 2d) + (l n - d) + Z n ,
and
S = In + (l n - d) + (Zn - 2d) + + (a + 2d) + (a + d) + a.
Adding the right sides, we have n terms each of which is a -I- l n .
Thus,
2S n = (a + Z n ) + (a + Z n ) + + (a + k) + (a + Z n ) = n(a + t).
Therefore,
(15-4) S n = (a + ^).
If we substitute a + (n - l)d from (15-3) for Z in (15-4), we
have another useful form for the sum. This is
(15-5) S n = [2a + (n -
Example 15-1. Determine which of the following sequences are arithmetic
progressions:
a) 3, 7, 10; 6) 6, 1, - 4; c) 3z - y, 4x + y, 5x + 3y.
Solution: a) Since the differences 10-7 and 7 3 are not equal, the sequence
3, 7, 10 is not an arithmetic progression.
b) In the second sequence, -4-1=1 6 = 5. Since these differences are
equal, the sequence 6, 1, - 4 is an arithmetic progression.
c) We find that (5x + 3y) - (4x + y) = x + 2y and (4s + y) - (3& - y)
= x + 2y. Since there is a common difference, the given sequence is an arithmetic
progression.
Example 15-2. Find the twelfth term, and also the sum of the first 12 terms,
of the arithmetic progression 4, 7, 10, .
262 Progressions Sec. 15-4
Solution: We have a = 4, n = 12, and d = 3. Then, by (15-3),
Z 12 = a + (n - l)d = 4 + (12 - 1)3 = 37.
Also, by (15-4),
Si. = \ (a + Z.) = y (4 + 37) = 246.
Example 15-3. The third term of an arithmetic progression is 3/4, and the
sixth term is 3/2. Find the twenty-second term.
Solution: By (15-3), h = a -f 2d and Z 6 = a -f 5d. Thus, we have
f a + 2d = 3/4,
{a + 5d = 3/2.
By solving these two linear equations, we find that a = 1/4 and d = 1/4. By
(15-3), l n = 1/4 + 21(1/4) = 11/2.
Example 15-4. Find each value of x for which the three quantities 3# 5, x + 4,
3x 2 form an arithmetic progression.
Solution: Applying the condition C B =B A, we have
(3x - 2) - (x + 4) = (x + 4) - (3x - 5).
o, i i. A - 15 m, -*u i- . . 25 31 37
Solving, we obtain ar = ~ The arithmetic progression is -j- > -7- > -j-
15-5. ARITHMETIC MEANS
The terms of an arithmetic progression between any two given
terms are called arithmetic means between the given terms.
If we let the given terms be a and Z n , any number, say fc, of means
may be inserted between a and l n by using the formula l n = a +
(n l)d with n = k + 2. As soon as we have found d, we can
insert the required means.
Example 15-5. Insert three arithmetic means between and 1.
Solution: Let a = 0, I* = 1, and n = 3 + 2 = 5. Then 1 = + 4d, and d =
113
Hence, the three means are j > ^ > j
If only a single arithmetic mean is to be inserted between two
given numbers, then the inserted value is called the arithmetic
mean of the given numbers. Thus, if a, x, b form an arithmetic
progression, x is called the arithmetic mean of a and 6. Since
6 x = x a,
Thus, the single arithmetic mean of two numbers is equal to one-
half their sum.
Sec. 1 5-5 Progressions 263
EXERCISE 15-2
In each of the problems from 1 to 9, determine if the given sequence is an arith-
metic progression. Find the next two terms of the extension of each arithmetic
progression.
1. 5, 8, 11, 14. 2. - 1, 7, 13, 19. 3. - 10, - 3, 4, 11.
4.4,12,19,27. 5- -H' 1 '*' -|'fi'-i'-Ig-
7. a, 6, 26 - a, 36 - 2a. 8. a + 6, a - 6, a - 26, a - 36. 9. ~-^ , a, ^^
^ ^
In each of the problems from 10 to 19, find l n and S n for the arithmetic progression.
10. 2, 8, 14, - to 12 terms. 11. 3, 6, 9, to 26 terms.
12. 22, 18, 14, - to 7 terms. 13. 1, 2, 3, to 10 terms.
14. 2, 4, 6, - to 50 terms. 15. 1, 3, 5, to 75 terms.
16. a, 2a, 3a, to 10 terms. 17. 0.2, 0.5, 0.8, - to 20 terms.
18. 1, 8, 15, to 35 terms. 19. 1, 2, 3, to 100 terms.
In each of the problems from 20 to 27, three quantities relating to an arithmetic
progression are given. Find the other two quantities.
20. a = 5, h = 36, n = 4. 21. a = 10, n = 10, d = 10.
22. S 2 i = 653, n = 21, a = 6. 23. n = 45, d = L 845 = 63.
i
24. Z 2l = 8, n = 21, d = \ 25. Z = 7, & = 52, d =
o ^
1 3 12*5
26. a = - i , d = | , = i-- 27. a = 1, S. = 45, d = 1.
28. Find the sum of the first 100 even integers.
29. Find the sum of the first n odd integers.
30. Insert five arithmetic means between 20 and 30.
31. Insert ten arithmetic means between 100 and 40.
32. Insert six arithmetic means between 3 and 2.
33. Find the arithmetic mean of 10 and 56 and that of 4 and 28.
34. Find the arithmetic mean of 28 and 65 and that of 33 and 78.
35. Insert k arithmetic means between - and a, where a ^ 0.
36. A display of cans in a grocery store is in the form of a pyramid whose base is an
equilateral triangle. If each side of the base contains 20 cans and the number of
cans decreases by one for each successive row, how many cans are in the display?
37. A lottery contains tickets numbered consecutively from 1 to 100. Customers
draw tickets and pay according to the number of the ticket, except that
tickets numbered above 50 cost just 50 cents each. How much money isi
collected if all tickets are sold?
38. Determine x so that x, x 2, 3x will be an arithmetic progression.
39. The sum of the first and fourth terms of an arithmetic progression is 20. The
sum of the third and twelfth terms is 36. Find the sum of the first 15 terms.
40. Find the sum of all multiples of 5 from 100 to 1,000, inclusive.
264 Progressions Sec. 1 5-6
15-6. HARMONIC PROGRESSIONS
A harmonic progression is a sequence of non-zero numbers whose
reciprocals form an arithmetic progression. Thus, a, b, c are in
harmonic progression if - > r > - form an arithmetic progression.
a o c
The terms of a harmonic progression between any two given
terms are harmonic means between the given terms.
To insert a desired number of harmonic means between two
numbers, we insert the same number of arithmetic means between
the reciprocals of the two given numbers and then invert the result-
ing terms.
The harmonic mean of two numbers is found in the follow-
ing manner. If a, x, b form a harmonic progression, then - > - > T
a x u
form an arithmetic progression, and - is the arithmetic mean of -
i x a
and r Hence,
o
Solution of this equation for x yields the harmonic mean
__ _2o6_
X ~ a + b'
Clearly the harmonic mean exists only if a + b = 0.
Example 15-6. Insert three harmonic means between 3 and 2.
Solution: The corresponding arithmetic progression is > Here a= $ >
o & o
Is = - > and n = 5. Hence, - = - - -f 4d, and d = ^1 '
It follows that the arithmetic progression is - > ~ o ' 77; ' TTT > o '
O O 1Z ZT: Z
24
Therefore, the three harmonic means are - 8, 12, -=-
EXERCISE 15-3
In each of the problems from 1 to 6, determine if the given sequence is a harmonic
progression. Find the next two terms of the extension of each harmonic progression.
,111 9 1 1 1 .111
L 3'7'U' 2 -4'8'l6* 8 '5'l6'l6'
*5'i'g- 5.16 f 8,f. ^ ^ .-2 f 2,|.
7. Find the tenth term of the harmonic progression - > = > > .
Z 7 U
8. Find the seventh term of the harmonic progression 6, 3, 2, .
9. Insert four harmonic means between 1 and 2.
Sec. 15-8 Progressions 265
10. Insert three harmonic means between ^ and 7
3 4
11. Insert four harmonic means between 6 and 24.
12. Find the harmonic mean of 6 and 9.
13. Find the harmonic mean of 24 and 72.
2 3
14. Insert nine harmonic means between - and ~
o 2t
15. If a 2 , 6 2 , r 2 form an arithmetic progression, show that 6-f-c, a+c, a+& form
a harmonic progression.
16. If a, 6, c form an arithmetic progression and 6, c, d form a harmonic progression,
show that ad = be.
17. If a? is the harmonic mean of a and b, show that 1 r = h T *
x a x b a b
18. If a, 6, c, d form a harmonic progression, show that 5 = _ ,
19. If a, 6, c form a harmonic progression, show that - = j-
C ~~" C
20. If the harmonic mean of a and 6 is equal to their arithmetic mean, show that
a = b y and conversely.
15-7. GEOMETRIC PROGRESSION
A geometric progression is a sequence of numbers in which each
term after the first is obtained from the preceding one by multi-
plying it by a fixed number; the multiplier is called the common
ratio.
The common ratio may be found by dividing any term by the one
immediately preceding it. Thus, 1, - > - > ^ is a geometric progression
Zi 4 O
in which the common ratio is --r-- = --f-- = -~-l=- Also,
o 4 4 ^ i Zi
\/2, 1, .= i - is a geometric progression in which the common ratio
V2 2
i s u 4= = 4= - 1 = i * V2 = -i
2 V2 \/2 A/2
It follows that a necessary and sufficient condition that three non-
C 1 R
zero numbers A, B, and C form a geometric progression is -^ = -j
15-8. THE GENERAL TERM OF A GEOMETRIC PROGRESSION
Let a denote the first term of a geometric progression, and let r
denote the common ratio. Then the progression may be written
as follows :
a, ar, ar 2 , ar 3 , , ar 71 " 1 .
Hence, if l n represents the value of the nth term,
(15-6) l n = ar*- 1
266 Progressions Sec. 1 5-9
15-9. SUM OF THE FIRST n TERMS OF A GEOMETRIC PROGRESSION
Let S n represent the sum of the first n terms of a geometric
progression. Then
S n = a + ar + ar 2 H h ar n ~ 2 + ar"- 1 .
Multiplying by r, we have
S n r = ar + ar 2 + ar 3 + + ar n ~ l + ar n .
Subtracting the first of these equations from the second, term by
term, we have
S n r S n = ar n a.
Therefore,
(15-7) S n = a ^ n j" 1 1) = a( * I /* (r ^ 1).
If we multiply both sides of (15-6) by r, we get rl n = ar n . Sub-
stituting in (15-7), we obtain another useful form for S H . This is
(15-8) S n = 5L=j (r * 1).
Note. If r = 1, these formulas do not apply; in this case, however,
the geometric progression becomes a + a H haton terms, and
S n = na.
Example 15-7. Determine which of the following sequences are geometric
progressions:
a) 2,6,18; 6) 5,10,30; c) *,^~-
y y
Solution: a) The ratios found by dividing each of the second and third terms by
the preceding one are 3 and 3. Hence, the sequence 2, 6, 18 is a geometric progression.
6) In this sequence, the ratios of consecutive terms are 2 and 3. Therefore, the
sequence 5, 10, 30 is not a geometric progression.
c) The given sequence is a geometric progression in which the common ratio
is x/y.
Example 15-8. Find h and S& in the geometric progression 6, 2/3, 2/27,
Solution: The common ratio is (2/3) * 6 = 1/9. Since a = 6 and n = 5, we have
\
o(l - r*) 6(1 - (1/9)*) _ V 59,049/ _ 14,762
6 1 - r 1-1/9 1-1/9 2187
Example 15-9. The fifth term of a geometric progression is 3, and the tenth
term is - 96. Find the common ratio and the first term.
Solution: By the formula (15-6) for the nth term of a geometric progression,
we have
ar 4 = 3 and ar 9 = - 96.
Sec. 15-10 Progressions 267
Dividing each side of the second equation by the corresponding member of the
first equation, we obtain r 5 = 32. Hence, r = - 2. Therefore, a( - 2) 4 = 3, or
Example 15-10. Find each value of x for which the three numbers x, x 2,
x + 1 form a geometric progression.
C B x 4- 1 x 2
Solution: If we apply the condition -5- = -r > we have - = = - Solving,
MJ A X ~~~ & X
we obtain z =4/5. Hence, the geometric progression is 4/5, - 6/5, 9/5.
15-10. GEOMETRIC MEANS
Terms of a geometric progression between any two given num-
bers are called geometric means between the given numbers. Let a
and l n be given numbers. Then k means may be inserted between
them by using the formula l n = ar"- 1 with n = k + 2.
Example 15-11. Insert three geometric means between 1 and 2.
Solution: Let a = 1, Z w = 2, and n = 5. Then / 5 = ar 4 and r = 2 1/4 . Hence,
the three means are 2 1 ' 4 , 2 1 ' 2 , 2 3 ' 4 .
If a, #, 6 form a geometric progression, then x is called a geo-
b x
metric mean of a and 6. Since - = -
# a
X = db
Hence, we note that a geometric mean of two numbers is the same
as a mean proportional between the two numbers.
EXERCISE 15-4
In each of the problems from 1 to 9, determine whether the given sequence is a
geometric progression. Find the next two terms of the extension of each geometric
progression.
1. 2, 8, 32. 2 .\ > - 1, 2. 3. 4, 16, 64.
&
4.2,4,6. 5.27,18,12. . * , J f -L
7 V3 V6 V3 . a 2 & Q 1 1 1
7 ' "T ' ~6~ ' T' 8> a ' 2" ' 3" 9. 1, - 1, 1.
In each of the problems from 10 to 15, find l n and S n in the given geometric
progression.
10. 4, 2, 1, - - ; n = 10., 11. 3, 2, 4/3, - ; n = 15.
12. 3, 9, 27, ; n = 45. 13. 100, - 10, 1, ; n = 101.
14. log 2, log 4, log 16, ; n = 10. 15. log 9, log 3, log V5, ; n = 6.
16. Find the sum of the first n terms of the geometric progression 1, =r j .
268 Progressions Sec. 15-10
17. For what values of x do x - 2, x 6, 2x + 3 form a geometric progression?
18. For what values of x do 3x + 4, x 2, 5x -f 1 form a geometric progression?
19. For what values of x is x -f 1 a geometric mean of 2x + 1 and a; 1?
20. Find a geometric mean of 4 and 16. Also, find their arithmetic mean.
21. Find a geometric mean and the arithmetic mean of 3 and 12.
22. Insert four geometric means between 1 and 32.
23. Insert five geometric means between 1 and 1,000,000.
24. Insert ten geometric means between 1 and 2.
25. If A, G, and H denote the arithmetic mean, a geometric mean, and the harmonic
mean, respectively, of two numbers a and 6, prove that G 2 = AH.
26. If the arithmetic mean of a and 6, when a, b 7* 0, is A, a geometric mean is G,
and the harmonic mean is H, find A G, G H, and A H.
2 4
27. Show that the sum of the first n terms of the geometric progression 1> o ' Q > * * '
is3(l (j 1 Discuss how this sum varies as n increases.
28. Show that the sum of the first n terms of the geometric progression 8, 4, 2, is
16(1 (sli* Discuss how this sum varies as n increases.
29. Find the sum of the first n terms of the sequence 1, 2x, 3x 2 , 4x 3 , . (Hint:
Let S n be the sum. Then compute S n xS n .)
f\ Q 11 Q<i i O
30. Find the sum of the terms of the finite sequence 2, = > =^ > =j > > =^
15-11. INFINITE GEOMETRIC PROGRESSION
A geometric progression in which the number of terms is infinite
is called an infinite geometric progression.
In Section 15-9, we found an expression for the nth partial sum
S n of a geometric progression. Hence, S n is the nth term of the
geometric series based on the given progression. Thus, we have
Si = a; 82 = a + ar; 83 = a + ar + ar 2 ; ; S n = a + ar + +
Furthermore,
Let us consider what happens to the sum of n terms of a geo-
metric progression when the number of terms increases indefinitely.
If a = 0, evidently S n = whether r ^ 1 or r = 1. In this case, the
number meets the requirement of a limit of S n , so that the sum S
of the infinite geometric series is equal to 0.
Now suppose a * 0. For this condition, we consider four cases.
Case 1. Assume that \r\ < 1. If r ^ 0, the numerical value of r
decreases as w increases. Moreover, by making the number of terms
Sec. 15-12 Progressions 269
sufficiently large, we can make \r n \ as small as we please. It follows
that if |r < 1, we can make S n differ from T-^_ by as little as we
please ; that is, S n approaches ^ as a limit. This condition may
be stated symbolically in the following manner:
(15-9) S = lim S n = - r ^ >
n-oo 1 r
where S is the sum of the infinite geometric series. Equation (15-9)
is true also if r = 0, since in this case S n has the constant value a.
Case 2. If r = 1, then S n na. Since a = 0, |S n | increases indefi-
nitely as n increases. Here { S n } diverges.
Case S. If |r| > 1, then jr 71 ] increases indefinitely as n increases.
Hence, so does | S n \ = | 7-^ ~^- | Again {S n \ diverges.
1 r 1 r
Case 4- If r 1, then the progression becomes, a, a, a, a, ,
( l)^ 1 a. If n is even, S n = 0. If n is odd, S n = a. In this case, we
say that S n oscillates between and a. Here also the series diverges.
The sum of an infinite geometric progression can therefore be
found by (15-9), but only when \r\ < 1. When r = 1, r 1, or
r\ > 1, the series diverges, and we say that the series has no sum.
For a further discussion of this topic, the student may refer to a
treatise on the theory of limits.
Example 15-12. The owner of a fleet of trucks finds that if used motor oil is
refined for re-use, 20 per cent of the oil is lost in the process. If he starts with 100
gallons of refined oil and re-refines this oil each time it becomes dirty, determine
the total amount of oil he has used before the entire 100 gallons is lost.
Solution: We begin with 100 gallons. After the first reclaiming operation, we
have 80 gallons of good oil. When this becomes used and is re-refined, we have 64
gallons; and so on. Theoretically, we would never use up the entire amount of oil.
However, the limit of the sum of the amounts of oil reclaimed is approximately
reached after a large number of operations. Hence, we have an infinite geometric
100
progression in which a = 100 and r = 4/5. Then S _ , = 500. Thus, by
re-refining the oil as it is used, the fleet owner has had the equivalent of 500 gallons
of oil.
15-12. REPEATING DECIMALS
JL
If a decimal contains a fixed sequence of digits which are
repeated indefinitely, we call it a repeating decimal. Thus,
0.135135 is a repeating decimal. This decimal is written 0.135,
the dots indicating the first and last digits of the sequence which is
to be repeated. Also, 0.34516 = 0.34516516516 . A repeating
decimal is wholly or partly an infinite geometric series. For
270 Progressions Sec. 1 5-1 2
example, since 0.34516 = 0.34 + .00516 + 0.00000516 + , it is
composed of the decimal 0.34 and an infinite geometric series in
which a = 0.00516 and r = 0.001. Hence, since |r| < 1,
' 00516 - a00516 - 34 172
' 0.999 ~~ 100 ' 33300 33300
Note that if we divide 172 by 33300, we obtain the repeating deci-
mal 0.00516.
Example 15-13. Express the repeating decimal 0.26*3 as an equivalent numerical
fraction.
Solution: We can write this decimal in the form
0.2 + 0.063 + 0.00063 + .
Hence, the required number consists of the decimal 0.2 plus an infinite geometric
series in which a = 0.063 and r = 0.01. The sum of the series, since \r\ < 1, is
0.063 0.063 7
1 - 0.01 ~~ 0.99 ~" 110 "
1 7 90
Therefore, 0.263=1+^=^.
EXERCISE 15-5
In each of the problems from 1 to 12, find the sum of the convergent series based
on the given infinite geometric progression.
1. | , - 1, | , - . 2. 1 , 1 , ~ , . 3. 3, V3, 1, - - - .
4. 1 , 1 , 1 , . 5. 8, 4, 2, ... 6. 8, - 4, 2, -...
.,24 fl 1 1 2 Qi 39
7. 1, g , g , . 8. ^ > 3 > g > * *' - 5 ' 25 ' " ' '
10. 0.5, 0.05, 0.005, .". 11. 0.18, 0.0018, 0.000018, ....
12. 0.3 + 0.012 + 0.00012 + 0.0000012 + .
13. Find the sum of the series based on the infinite geometric progression 24,
8, 5 > . Also, find the sum of the first 20 terms of this progression and com-
o
pute the error introduced by using S instead of $20-
14. What would be the error if S were used instead of S n for the sum of the geometric
progression 48, - 36, 27, - to 10 terms?
15. A ball is dropped from a height of 3 feet. On each rebound it bounces back to
three-fourths the height from which it last fell. Assuming that this bouncing
continues indefinitely, find the distance it travels in coming to rest. How far
has it traveled after bouncing ten times?
16. A swinging pendulum will gradually come to rest as a result of friction. If, on
each upswing, the pendulum swings through 98 per cent of the arc through
which it fell, and if the initial arc for one complete swing was 20 inches, find the
distance traveled before the pendulum comes to rest.
Sec. 15-13 Progressions 271
Convert each of the following repeating decimals to fractional form.
17. 0.1. 18. O.i5. 19. 0.90. 20. 0,243.
21. 0.16. 22. 0.142857. 23. 2.9. 24. 1.1234.
25. 0.11542. 26. 2.123.
15-13. THE BINOMIAL SERIES
We shall now consider the binomial expansion when n is any real
number. If we let a = 1 and b = x in (4-13), the binomial formula
becomes
/ie IA\ /t i \ 11 i n(n L- 1) o , n(n 1) (n 2)
(15-10) (1 + x) n = 1 + nx + ^ 2T - x 2 + ~ x 3
n(n - 1) (n - 2) (n - r + 1)
+ ...+ - x +....
The right member of (15-10) is called a binomial series.
We saw in Section 4-6 that, if n is a positive integer, the series
on the right in (15-10) terminates with x n , and (15-10) is true.
Otherwise, the terms of the series continue indefinitely, giving rise
to an infinite series.
The question which now arises is whether (15-10) is valid when
n is not a positive integer ; that is, whether the series on the right
converges, and, if so, whether its sum is equal to (1 + x) n . It is
proved in the study of series that (15-10) is indeed valid if \x\ < 1.
It follows that, if \x\ < 1, we may obtain the value of (1 + x) n as
accurately as we please by taking sufficiently many terms of the
series in (15-10).
The expansion can readily be extended to (a + 6) n when n is not
a positive integer. In this case, the expression may be written as
follows :
(15-11) (a + &) = fo(l + -Yr = a" fl + -T-
L. \ a/ j L. d-j
Here x = - > and the expansion of (a + 6) n is valid if - < 1.
a \ a \
Example 15-14. Find the first five terms of the expansion of (1 + z)~ 3 it\x\ < 1.
Solution: By (15-10),
(-8)(-4)(-6) ja (-8) (-4) (-5) (-
3! 4!
272 Progressions Sec. 15-13
Example 15-15. Find ^L04.
Solution: j/TM = (1 + 0.04) "3. Hence, n = 1/3 and x = 0.04, and the ex-
pansion is 5
(1 + 0.04)3 = 1+| (0.04) + 2! (0.04)2 + 3 ^ 3 ^ 3 ^ (0.04)3+ ....
The approximate value of the sum of this series is
1 + 0.013333 - 0.000177 + 0.000004 = 1.013160.
We have here a case of an alternating series, that is, a series in which the terms
are alternately positive and negative. It is proved in the study of series that, if in
an alternating series each term is numerically less than the preceding term and
lim a n := 0, the error introduced by using S n as the sum of the series is
numerically less than the value of the first term omitted; that is, \S n - S\ < \a n +i\.
If in the present example we take for S the value S 3 = 1 + 0.013333 - 0.000177
= 1.013156, the error in so doing is less than the value of the fourth term 0.000004.
Example 15-16. Find the first four terms of the expansion of .,
F
- x*
Solution: First we apply (15-11) to convert the given expression to a suitable
form, as follows:
1 / / r2\\-i/3
-^ = (8 -.)-. = (8(1-!))
(/v2\-l/3 1 / 2\-l/3
^T) =K 1 -!-)
Hcnco, by (15-10), if x* < 8,
_i ,
~
a 2 \ (-1/3) (-4/8) / x*
"*"
i \ s/"" 1-2 V
(-1/3) (-4/3) (-7/3)7 x*\* \
"*" 1-2-3 V 8/" 1 ""/
24 288 "" 20,736
EXERCISE 15-6
In each of the problems from 1 to 10, find the first four terms of a binomial
expansion of the given expression.
i. ^ 2 - T- 3 - vrr^. 4. vr^. 5. (
1 ~T" X 1 X
6. (1 - z)-i/'. 7. -4- 8. x _? v ' 9. 7 ^-r-- 10. (1 +ir) l/aj .
x + y (x + i/) 2 (x + 2/) 3 v -r /
Find the approximate value of each of the following numbers by means of a
binomial expansion, using four terms of the expression.
11. (1.02)io. 12. (1.01) 13 . 13. (1.04) 8 . 14. (l.l) 10 . 15. (0.98) 8 .
16. 49 4 . 17. (0.99)'. 18 . 51 3. 19 . (1.03)1/2. 20. (0.97)~ 2 .
16
Mathematical Induction
16-1. METHOD OF MATHEMATICAL INDUCTION
When a certain type of formula or proposition has been verified
in specific cases but is not known to be true in general, the method
of mathematical induction is often found extremely valuable in
determining its validity.
Suppose that a statement involving a positive integer n is to be
proved true for all values of n greater than or equal to a particular
initial value. We begin by showing the result to be true for the first
value of n. We then assume that k is some particular integral value
of n for which the statement holds. With this assumption as a
basis, we establish the validity of the statement in the next suc-
ceeding case, namely, that in which n k 4- 1. In other words, we
prove that if the statement is true for any specific integral value
of n, say n = k, then it is also true for the next larger value of n,
namely, n = k 4- 1. Suppose for example, that 1 is the initial value
of n. Then the second step establishes that if the statement is
true for n = l, it is also true for n = 1 + 1, or 2 ; if it is true for
n = 2, it is also true for ft = 2 + 1, or 3 ; and so on. As a consequence
of this, we conclude that the statement is true for all values of n
greater than or equal to the initial value, here 1. A proof by
mathematical induction, therefore, consists of two parts and a
conclusion.
Part 1. Verification that the statement is true for some initial,
value of n, generally n = 1. (This initial value is the smallest value
of n for which the statement is to be proved true.)
Part 2. Proof that whenever the statement is true for some par-
ticular value of n, say for n = k, then it is true for the next larger
value of n, that is, for n = k + 1.
273
274 Mdfhemafi'ca/ Induction Sec. 16-1
Conclusion. If both parts of the proof have been given, then the
statement is true for all positive integral values of n greater than
or equal to the one for which the verification was made in Part 1.
The reasoning process involved here, which consists in taking
an initial integer and then repeatedly taking successors, can be
exemplified in terms of climbing a ladder. Part 1 puts us on the
bottom rung of a ladder. Part 2 shows us how to get from any rung
we have reached to the next higher rung. The conclusion states
that if we know how to get on the bottom rung of the ladder, and if
we know how to get from any rung to the next higher one, then we
can reach all rungs, and hence can climb the ladder.
The following examples will illustrate the method.
Example 16-L If n is any positive integer, prove that
(16-1) 1+1+1 +.
1-22-33-4 ^ n(n + 1) n + 1
Solution:
Part 1: The formula is true when n = 1, since
1 = 1 1=1.
1-2 1+1* r 2 2*
Part 2: Let k represent any particular value of n for which (16-1) is true. Then
I I I I - -
l-2" 1 "2-3" t "3-4" t " "*" fc(fc + 1) ~~ fc + 1
We now wish to prove that (16-1) is true also for the next larger value of n,
namely, n = k + 1. The sum on the left in (16-1) when n = k + 1 can be obtained
by adding its last term, which is 7, . iwy r-m > to both sides of (16-2). Hence,
we have (* + 1) (* + 2)
/LL+JL+JL + .
VI -2 "^2-3 ^3-4 T
k(k + !)/ ^ (k + l)(k +2)
k .
k + 1 ^ (* + 1) (k + 2)
k 1 _fc 2 +2fc+l_fc+l_ k + l
But, since fc + j + (t + D (t + 2 ) ~ (fc + i) (fc + 2) ~ t + 2 ~ (* + 1) + I 1
we obtain
, lft ox ! . 1 . ! . 1 _ k + l
(16-3) TTo + oTo + oTT +
1 2 ^ 2 3 ^ 3 4 ^ ^ (A? + 1) (k + 2) " (* + 1) + 1
The members of (16-3) are the same as those of (16-1) when n = k + 1. Hence,
we have shown that (16-3) is true if (16-2) is true, in other words, (16-1) is true
for n = k + 1 if it is true for n = k.
Conclusion: We have shown by verification that (16-1) is true when n = 1.
Therefore, since (16-1) is true for n = 1, it follows from Part 2 that (16-1) is true
for every positive integer n.
Sec. 16-2 Mathematical Induction 275
Example 16-2. Prove that (x - y) is a factor of (x n y n ) if n is any positive
integer.
Solution:
Part l:Itn = 1, then x n y n = x ?/, which is seen to have (a; y) as a factor.
Part 2: Let A: be a specific value of n for which (x n 7/ n ) has (x - y) as a factor.
We shall now prove that if (x y) is a factor of (x k ?/*), it is also a factor of
(a*M - y*+i). We have
x *+i _ yk+i = x *+i _ 33,* 4. X2/ * - y *+i = x ( x k _ y *) 4. <,*( _ y).
By assumption, (x - y) is a factor of (x* - y*). Also, by inspection, (x y) is a
factor of y k (x - y). Hence, (x - y) is a factor of the left member (x k+l - y** 1 ).
Therefore, if the conclusion is true for n = , it is also true for n = k + 1.
Conclusion: By Part 1 of the proof, (# - y) is a factor of (x n - y n ) when n = 1.
Therefore, by virtue of Part 2, the desired conclusion is true for any positive
integer n.
16-2. PROOF OF THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS
We shall now prove that the binomial formula,
(16-4) (a + b) n =a n + na n ~ l b + n ^ n T ^ an " 2fc2
...
(r - 1)1
n(n-l)...(n-r+l) an _ r6r + . . . + ^
r!
is true for every positive integral value of n.
Proof. Part 1. When n = 1, each side of (16-4) becomes a + 6.
Hence, (16-4) is true for n = 1.
Par #. Let fc be any specific value of n for which (16-4) is true.
Then we have
(a + 6)* = a* + ka k ~ l b + k ^ k ~ X) a*~ 2 6 2
1 >
Multiplying each member of this equation by (a + &), we obtain
' ' (k "" r + 1
(a + 6)*+i = a** 1
H ---- + ab k + a k b -\
276 Mathematical Induction See. 16-2
Hence, by combining terms, we get
(o + 6)* +1 = a** 1 + (Jfc + l)a6 + . . .
In this result the sum of the coefficients of a k ~ r+l b r is obtained as
follows :
k(k - 1) (k - r + 2) (k - r + 1) fc(Jfc - 1) (fc - r + 2)
r! + (r-1)!
fc-r+1
r
~L
r
_ (fc + 1) (fc) (fc - 1) (k - r + 2)
r!
We note that the value of (a + 6)** 1 , obtained as the product of
(a + b) k and (a + 6), is exactly the same as the expansion which
would be obtained from (16-4) with n = fc + l. Hence, we have
shown that if the binomial formula holds for n = k, it must hold for
n = fc + l.
Conclusion. The binomial formula was seen to be true for n = 1.
Therefore, by virtue of Part 2, we may conclude that it is true for
every positive integer n.
EXERCISE 16-1
Prove by mathematical induction that each of the statements in Problems 1 to
20 is true for all positive integral values of n.
.
2. 1 + 3 4- 5 H ---- + (2n - 1) = n 2 .
3. 2 + 4 + 6 H ---- + 2n = n(n + 1).
5.3+6
6. 4 + 8 4- 12 H ---- + 4n = 2n(n -f 1).
7 . 1+4+7 + .. . +(3W _2)
9.
10.
Sec. 16-2 Mathematical Induction 277
12. I 8 + 2 s + 3 8 +
13. I 3 + 3 3 + 5 3 + + (2n - I) 3 = rc 2 (2r* 2 - 1).
14. 2 3 + 4 8 + 6 8 + + (2n) 8 = 2n 2 (n + I) 2 .
15. 2 + 2 2 + 2 3 + +2 = 2(2" - 1).
16. 3 + 3 2 + 3 3 + + 3 = | (3 - 1).
&
17. 4 + 4 2 + 4 3 + - + 4 = | (4 -,,1).
18. I 2 + 3 2 + 5* + 7* + +frt -% = n(2n -
19 ^
^'
1 3 3 5 " 5 7 "" " (2n - 1) (2n + 1) "" 2n + 1
2ft 1 i 1 i 1 i i 1 = n(n+3) ^
1 2 3 ^ 2 3 4 "*" 3 4 5 "*" "^ n(n + 1) (n + 2) 4(n + 1} (n + 2)
21. Prove that x 2n - 2/ 2n is divisible by x + y for every positive integer n.
22. Prove that x 2n ~ l + y*n-i {$ divisible by x -f y, for every positive integer n.
23. By using mathematical induction, prove the formula for the sum of an arith-
metic progression.
24. By using mathematical induction, prove the formula for the sum of a geometric
progression.
17
Permutations, Combinations,
and Probability
17-1. FUNDAMENTAL PRINCIPLE
We begin the study of permutations and combinations by con-
sidering the following principle, which is fundamental for the
entire subject.
Fundamental Principle. If one thing can be done in a ways, and
if, for each such way, a second thing can be done in b ways, then
the two together can be done in a b ways.
To understand why the principle is true, note that for each of the
a ways of doing the first thing, there are b ways of doing the
second ; hence, both the first and the second things taken together
can be done in a b ways.
The following examples will illustrate the reasoning upon which
the principle is based, as well as an obvious extension of the prin-
ciple to the case when more than two things are to be done.
Example 17-1. In how many ways can two officers, a chairman and a secretary,
be selected from a committee of five men?
Solution: By the fundamental principle, the problem is equivalent to determining
the number of ways in which the two things can be done together. The first
position can be filled in 5 ways; that is, there are a = 5 ways of selecting a chairman.
For each of these possible selections, there are 6 = 4 ways of filling the position of
secretary from the remaining men. Hence, the number of ways of selecting a
chairman and secretary is a 6 = 5 4 = 20.
Example 17-2. How many three-digit numbers can be formed from the ten
digits 0, 1, 2, , 9, if a) repetitions of digits are not permitted; 6) repetitions
are permitted?
Solution: a) Here we have three things to do or places to fill. The hundreds
place can be filled in 9 ways, since must be excluded from this place. The tens
278
Sec. 17-2 Permufaf/ons, Comb/naf/ons, oncf Probability 279
place can then be filled in 9 ways from any of the remaining 9 digits. Finally, the
units place can be filled from any of the remaining 8 digits. There are, therefore,
9.9.3 = 648 three-digit numbers in which no two digits are alike.
b) If repetitions are permitted, there are 9 ways to fill the hundreds place and
10 ways to fill each of the tens and units places. Hence, there are 9 10 10 = 900
three-digit numbers.
EXERCISE 17-1
1. Nine persons apply for each of two vacant apartments. In how many possible
ways can both apartments be rented?
2. A large room has eight doors. In how many ways can a person enter the room
by one door and leave by a different door?
3. If three dice are thrown, in how many ways can they fall?
4. There are eight men and six women in a club. In how many ways can two
officers be selected so that one is a man and one is a woman?
5. How many possible four-digit numbers are there in a telephone exchange which
uses only four-digit numbers and excludes 0000?
6. How many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9?
How many of these are larger than 700,000?
7. How many numbers of six different digits can be formed from the digits 2, 3, 4,
5, 6, 7, 8, 9? How many of these are larger than 700,000?
8. A woman has seven guests at a party. If she chooses her seat first, in how many
ways can she seat her guests?
17-2. PERMUTATIONS
An ordered arrangement of all or any part of a set of things is
called a permutation. Specifically, suppose we have n distinct things
and wish to select r of these to be arranged in a definite order.
Each such ordered arrangement is called a permutation of n things
r at a time. The number of all such permutations is denoted by
n P r . Thus, 5 P 2 is read "the number of permutations of 5 different
things 2 at a time."
In Example 17-1, the possible number of ways of selecting a
chairman and secretary from a committee of five men was found
by means of the fundamental principle to be 5 4 = 20. This is pre-
cisely equal to 5 P 2 , since it is the number of ways in which two men
can be chosen from among the given five men and arranged in the
two offices. The number of permutations of n things r at a time is
given by the formula
(17-1) n Pr = n(n - 1) (n - 2) (n - r + 1).
The truth of (17-1) is readily shown as follows. The first of the
r places can be filled in n ways. Then the second can be filled in
(n 1) ways, the third in (n 2) ways, and so on. In general, the
280 Permutations, Combinations, and Probability Sec. 17-2
number of ways of filling each place is n minus the number of places
already filled. Therefore, when the rth object is chosen, (r 1)
places have already been filled, and the rth place can then be filled
in n (r 1) , or n r + 1, ways.
In particular, if r = n, the last factor becomes n n + 1 = 1. We
then have
(17-2) n P n = n(n - 1) (n - 2) 1 = nl
This formula gives the number of permutations of n different
things taken all at a time.
If both the numerator and the (understood unit) denominator
of (17-1) are multiplied by (n r) !, we obtain the following alter-
nate formula :
H7 * P _ n(n - 1) (n - 2) (n - r + 1) (n - r)l _ n\
(17-6) n r r - ( n -r)! ~ (n - r) ! '
Here it is agreed that by definition ! = 1. Hence, (17-3) holds for
r ~n.
For example, by (17-2),
8 P 6 = n(n - 1) (n - 2) (n - r + 1) = 8 7 6 5 4 = 6720.
Using (17-3), we have also
_ nl _8!_8-7-6*5-4-3-2.1
aft ~ (^=7)1 " 8! T2^ 672 '
17-3. PERMUTATIONS OF n THINGS NOT ALL DIFFERENT
Suppose it is required to find the number of indistinguishable
distinct permutations of n things all at a time, if n\ things are
regarded as indistinguishable, n^ other things are regarded as indis-
tinguishable, and so on. Let us consider, for example, the number
of permutations of the letters a, a, a, b taken four at a time. For
convenience, indistinguishable objects are given the same notation.
Denote by P the desired number of distinct permutations. Evi-
dently, P is less than ^P^ which is the number of permutations that
could be effected if all the letters were distinguishable. For in any
one of the P permutations, say (a b a a) , any rearrangement of the
a's among themselves would not change the permutation. If, how-
ever, we assigned subscripts to a in this permutation, as in
(eh 6 02 da), we could permute these three distinct letters among
themselves in 3 ! ways. This can be done for each of the P permu-
tations of the letters a, a, a, 6. We would then obtain P 3 ! per-
mutations of the four distinct letters a l9 o^, a 3 , b taken four at a
time. There are therefore 4 ! permutations altogether. Hence,
P- 3! =4!, or P = Jj = 4!
3! ~3!1!
Sec. 17-4 Permutations, Combinations, and Probability 281
In general, the number of distinct permutations of n things taken
all at a time, if HI things are alike, n 2 other things are alike, n 3 other
things are also alike, and so on, equals
n\
(17-4) P =
. COMBINATIONS
n\\
A combination is a set of all or any part of a collection of objects
without regard to the order of the objects in the set. We use the
symbol n C r to represent the total number of all combinations of n
different things taken r at a time.
The different sets of the four letters a, 6, c, d taken three at a
time, without reference to the order in which the letters are
arranged, are (abc) 9 (abd), (acd), (bed). From each of these four
combinations, we can form 3 !, or 6, different permutations of the
four letters taken three at a time. For example, from the combina-
tion (abc) we can form the distinct permutations (abc), (acb),
(bac), (bca), (cab), (cba). Hence, each of the four combinations
contributes 3 P 3 = 3 ! permutations to the total number of permu-
tations. Thus, there are 4 C 3 combinations of the four letters if we
disregard order, and there are 3! ordered arrangements or per-
mutations of each combination. Hence, we have J3 = 3 ! 4^3, or
4 C 3 = 5 = 4>3 >2 ' l = 4 combinations. This is the same as the
o ! 1 2 o
number we obtained at the beginning of the paragraph.
In general, if we divide the total number of permutations of n
things r at a time, or n P r , by the number of permutations, or r!,
contributed by each combination, we obtain the total number of
combinations. Symbolically, we have
n4 r = ? * rbt>
or p
(17-5) C ' = 7T'
If we note that r I = r P r , then we can write the following inter-
esting relationship :
C n * r
.. r ~ ~~p~
r r r
Replacing n P r by its equivalent expression from (17-1) or (17-3)
we have
__ n(n - 1) * (n ~ r + 1) _ ' n\
282 Permutations, Combinations, and Probability Sec. 17-4
-r) in
n C n _ r =
If r is replaced by (n r) in (17-6) , we obtain
n\
(n- r)!r!
Hence,
(17-7) nC r = nCn-r-
Afote. Having agreed that ! - 1, we can supply (17-1) or (17-3)
to permutations or combinations of n things zero at a time. Con-
sidering (17-5), we have
(17-8) n Co = 0!(n l 0)I = o! = L
This result agrees with the intuitive conclusion that there is only
one such "empty" combination. Similarly, n P = 1.
17-5. BINOMIAL COEFFICIENTS
By referring to the development of the binomial formula in Sec-
tion 4-6, we note that the expansion of (a + b) n involves the prod-
uct (a + 6) (a + 6) (a + 6) taken without regard to order. This
fact suggests the use of combinations in the coefficients of the
expansion.
We said in Section 4-7 that the coefficient of the term involving
a n-r b r is " ' " - . This j g precisely the f or mula for
r!
the combination of n things r at a time, or n C r , obtained in Section
17-4. The binomial formula may therefore be written as follows :
(17-9) (a+b)* = n Coa+nCia n -ib+nC 2 a n - 2 b 2 +' .+#'"&+ -+ n C n b\
For example, the expansion of (a + 6) 4 takes the following form :
This reduces to
a 4 + 4a 3 6 + 6a 2 6 2
If we let a = 6 = 1 in the expansion for (a + b) n in (17-9), we
obtain
(1 + 1) = n C + nCl + n C 2 + - + nC n .
Since n C = 1, Ci + C 2 + + n C n = 2 - 1.
Thus, the total number o;f combinations of n things taken succes-
sively 1, 2, , ft at a time is 2 n 1.
EXERCISE 17-2
1. Evaluate 5^2, 7^3, 12^6, 21^4,
2. Evaluate 10^4, nCa, looC's, 21^5, 100^95*
Sec. 17-6 Permutations, Combinations, and Probability 283
3. Form all possible distinct permutations of the letters of the word theory.
a. How many are there? b. How many begin and end with a vowel? c. How
many begin or end with a vowel?
4. How many distinct permutations are there of the five letters a, 6, c, d, e taken
three at a time? Write them out.
5. In how many different ways can a dime, a quarter, and a half-dollar be dis-
tributed among five boys?
6. How many different sums can be formed with a penny, a nickel, a dime, and
a quarter?
7. Expand (a -f 6) 8 , using combination symbols.
8. From the six digits 1, 2, 3, 4, 5, 6, form all permutations taken five at a time.
a. How many are formed? b. How many begin with 4? c. In how many does
the digit 3 not appear?
9. How many distinct permutations can be made from the letters of the word
probability taken all at a time?
10. How many different combinations are there of 4 identical nickels ; 5 identical
dimes, and 6 identical quarters?
11. How many different straight lines are determined by twelve points, no three
of which are in a straight line?
12. In how many different ways can a student select seven questions out of ten
on a test?
13. How many different weights can be formed with six objects weighing 1, 2, 4, 8,
16, and 32 pounds, respectively?
14. In how many different ways can signals be made with seven different flags,
where a signal is a set of one or more flags arranged in a specific order?
15. In how many different ways can the 52 cards of a bridge deck be dealt among
four players?
17-6. MATHEMATICAL PROBABILITY
If, in a given trial, an event can happen in h different ways and
can fail to happen in / ways, and if all the h + / ways are equally
likely, then the probability p that the event will happen in this
trial is
(17-10) p =
The probability q that the event will fail to happen is
(17-11) g-
Note that O^p^ 1, O^g^l, and p + q = 1. Two illustrations of
probability follow.
If a bag contains 3 green marbles and 4 yellow marbles, all
exactly alike except for color, the probability of drawing a single
green marble is
3 3
284 Permufaf/ons, Comfamaffons, and Probability Sec. 17-6
A die can fall in six ways. The probability of getting 5 or more
2 1
ith one throw of a die is - > or - > since th
o o
either a 5 or a 6, for the event to happen.
2 1
with one throw of a die is - > or - > since there are two possible ways,
o o
17-7. MOST PROBABLE NUMBER AND MATHEMATICAL EXPECTATION
Let p be the probability of occurrence of some event. Further-
more, suppose that n trials of the event are made, of which h are
successful. Then - is called the relative frequency of success for
U h
the trials which occurred. It is not to be expected that - = p. It is
TV
shown in more advanced treatments of probability, however, that if
n is large, it is very likely that the relative frequency is approxi-
mately equal to p. Also, the larger we take n, the more likely it is
that - approximates p closely. Moreover, it can be shown that the
id
most probable or expected number of occurrences of the event for n
trials is up.
For example, when a coin is tossed, the probability of getting a
head is 1/2. In 1,000 trials the expected number of heads is there-
fore 500. This does not mean, however, that if the first 100 trials
result in 75 heads and 25 tails, we should expect 25 heads and 75
tails in the next 100 trials. Actually, since one toss of the coin does
not affect the next one, we should expect about 50 heads and 50 tails
in the next 100 trials. Moreover, we may expect about 450 heads
and 450 tails in the next 900 trials.
If p is the probability of winning a certain amount of money in
case a certain event occurs and m is the amount of money to be won,
the mathematical expectation is defined to be pm. For instance, if a
person can win $12 provided he throws an ace with a die, his expec-
tation is - ($12) = $2. Hence, $2 is the fair amount he should be
willing to pay to make the trial.
17-8. STATISTICAL, OR EMPIRICAL, PROBABILITY
It is frequently impossible to have sufficient knowledge before-
hand of all the conditions that might cause an event to happen or
fail to happen. In such a case, however, it may be possible to deter-
mine the relative frequency of the occurrence of the event from a
large number of trials. Thus, if an event has been observed to
happen h times in n trials, and n is a large number, then until addi-
Sec. 17-$ Permufaf/ons, Combinations, and Probability
tional knowledge is available, we define the statistical probability,
or empirical probability, to be
(17-10 P = l>
where - is the relative frequency.
n
Example 17-3. A molding machine turns out 12 parts per minute. Inspection
experience has shown that there are 20 defective parts per hour. What is the
probability that a single part, picked at random, will be defective? In a run of
10,000 parts, how many defectives should be expected?
Solution: The parts are produced at the rate of 720 per hour, and 20 of them are
defective. Hence, the probability that a single part selected at random will be
defective is =7^ = In a run of 10,000 parts, we should expect ^ (10,000), or
(ZO ob oO
approximately 278, defective parts.
17-9. MUTUALLY EXCLUSIVE EVENTS
Two or more events are mutually exclusive if not more than one
of them can happen in a given trial. The following theorem may
be stated.
Theorem. The probability that some one of a set of mutually
exclusive events will happen in a given trial is the sum of the indi-
vidual-event probabilities.
Proof. Consider, for simplicity, a set of two mutually exclusive
events. Suppose that the first can happen in hi ways and the second
can happen in h 2 ways, and let n be the total number of ways in
which the two events can happen or fail to happen. Then p\ =
and p2 = are the corresponding probabilities of the two events.
it
Since the n events are mutually exclusive, the hi ways are differ-
ent from the h 2 ways, and the number of ways that either the first
or the second event can happen is therefore hi + h%. Hence, the
probability p that either the one event or the other will happen is
(17-12) p.idt*. .,.&.,..
For example, suppose that a bag contains 2 green marbles, 3 yel-
low marbles and 5 brown marbles. If a marble is drawn at random,
2
the probability that it is green is ^ > and the probability that it is
o
yellow is Hence if a marble is drawn, the probability that it is
23 51
either green or yellow is + ^ > or ^ > or ^
1U lu lu ^
286 Permutations, Combinations, and Probability Sec. 1710
17-10. DEPENDENT AND INDEPENDENT EVENTS
In case two or more events are not mutually exclusive, they are
dependent if the occurrence of any one affects the occurrence of the
others, and they are independent if the occurrence of one does not
affect the occurrence of the others.
For example, if a card is drawn from a deck of 52 cards and the
card is not replaced before a second is drawn, then the second
drawing is dependent on the outcome of the first. If, however, the
first card is replaced, then the second drawing is independent of
the first. In the latter case the two drawings are equivalent to
simultaneous drawings from two decks.
We shall now state and prove the following theorem relating to
dependent and independent events.
Theorem. The probability that two dependent or independent
events will occur (successively if dependent; successively or simul-
taneously if independent) is the product of their individual
probabilities.
Proof. Suppose that the first event can happen in hi out of a
total of ni different ways, and that the second event can happen
in ft out of n^ different ways. Then it follows, by the fundamental
principle in Section 17-1, that the two events can happen together
in hih^ ways out of a total of n^n 2 different ways. Therefore, the
probability that both events will happen is
(17-13) P
' * ni H2
Example 17-4. Two cards are drawn from a deck containing 52 cards. Find the
probability that both cards are aces when the first card is not replaced before the
second is drawn.
Solution: We shall begin with a listing of the following useful probabilities:
4
1) The probability of drawing an ace from a deck of 52 cards is ~
oZ
2) If the first card is an ace and it is not replaced, the probability of drawing
another ace is -^
01
3) If the first card is not an ace and is not replaced, the probability of the
second being an ace is
51
4
4) If the first card is replaced, the probability of the second being an ace is -r^
This drawing is entirely independent of the first drawing.
Consider, now, the given problem. The probability that the first card is an ace is
4 1
Pi = - = If the first card drawn is an ace, then the probability that the
O/ 16
Sec. 1 7-1 1 Permufat/ons, Comb/naf/ons, cmcf Probability 287
3 1
second card drawn is an ace is p2 = ^r = T Hence, the probability that both
oi ii
will be aces is pl p a =-._ = .
17-11. REPEATED TRIALS
Theorem. If p is the probability that an event will happen in any
trial, and q = 1 - p is the probability that it will fail, then the prob-
ability that it will happen exactly r times out of n trials is
(17-14) nC,tr<r* = n i W- r -
Proof. The happening of the event in exactly r trials and its fail-
ure in the remaining n r times are independent events. Hence,
the probability, by the theorem in Section 17-10, is p r q n ~ r . But these
r trials can be chosen from the n trials in n C r ways. Since these
ways are mutually exclusive, the total probability is n C r p r q n ~ r * Note
that this expression is the (r + l)th term of the binomial formula
for (q + p) M , since
(q + pY = q n + n Ciq n ~ l p + n C 2 q n ~ 2 p 2 + + n C r q n - r p r + + p n .
The successive terms of this expansion give the probabilities that
the event will happen exactly 0, 1, 2, , r, n times in n trials.
An event will happen at least r times in a given number of n
trials if it happens n, n 1, , r + 1, or r times. Since these events
are mutually exclusive, the probability that an event will happen at
le&st r times is given by the sum
Example 17-5. What is the probability of tossing an ace exactly three times in
four trials with one die?
Solution: Since the probability of tossing an ace in one trial is ^ and the prob-
PJ
ability of failure is ^ * we may substitute in the term n V<r~ r P r of the binomial
formula. The result is
* (f )'(!)'= '(I) (s> = sir
Example 17-6. What is the probability of tossing an ace at least twice in four
trials with one die?
Solution: The event will happen at least twice if it happens 4, 3, or 2 times.
Hence, the probability is given by the following sum :
- '(!)'=
288 Permufaf/ons, Comfa/naf/ons, and Probability Sec. 17-1 1
EXERCISE 17-3
1. A certain event can happen in four ways and can fail to happen in six ways.
What is the probability that it will happen? If $60 can be won on the event,
what is the mathematical expectation?
2. A box contains 5 white balls, 4 red balls, and 13 black balls, a. If one ball is
drawn out, what is the probability that it is red? b. What is the probability
that it is white or red?
3. a. If one die is thrown, what is the probability that a "1" or a "2" will turn up?
b. What is the probability that a "3" or larger number will turn up?
4. When a coin was tossed 100 times, 80 heads and 20 tails turned up. If the
tossing were continued until 200 tosses had been made, what would be the
most probable number of tails in the second 100 tosses?
5. A bag contains five $1 bills, ten $5 bills, and twenty $10 bills. If one bill is
drawn, what is the mathematical expectation?
6. What is the probability of throwing a "7" or an "11" on one throw of two dice?
7. An automobile owner carries $1,000 theft insurance on his car. If, during the
past year, 237 out of 97,864 automobiles registered in his area were stolen,
what is the mathematical value of the policy?
8. In a city of 77,000 families, a careful sample of 800 families showed that 120
of the sample families owned their own homes, a. What is the probability
that a family selected at random in the city owned its home? b. What is the
expected number of families in the city who own their own homes?
9. In a certain city 28,600 persons voted for one candidate for an office, and
23,100 voted for his opponent. What is the probability that a voter chosen at
random voted for the winner?
10. A bag contains 5 red balls and 9 black balls. If two balls are drawn in suc-
cession, and the first is not replaced, find the probability that the first is red
and the second is black.
11. In a baseball tournament the probability that team A will win is = > and the
I '
probability that team B will win is ^ Find the probability that one of these
two teams will win.
2
12. The probability that team A will reach the finals of a tournament is = >
I '
and the probability that it will win the finals is - Find the probability that
team A will win the tournament.
13. Find the probability of throwing three successive fours on a pair of dice.
14. The probability of A winning a game when he plays it is j He is scheduled
to play four times, a. Find the probability that he will win exactly three times.
b. Find the probability that he will win at least three times.
15. Three dice are tossed, a. Find the probability that exactly two threes will
turn up. b. What is the probability that at most two threes will turn up?
18
Solution of the
General Triangle
18-1. CLASSES OF PROBLEMS
There are certain relationships among the lengths of the sides
and the trigonometric functions of the angles of every triangle. If
one side and any two other parts of a triangle are given, the
remaining parts can be determined; that is, the triangle can be
solved. The three given parts may comprise any one of the follow-
ing four combinations :
Case I. One side and two angles.
Case II. Two sides and the angle opposite one of them.
Case III. Two sides and the included angle.
Case IV. Three sides.
In this chapter, we shall discuss methods for treating these four
cases. For convenience, we shall let ABC denote any triangle
whose angles are A, B, and (7; and we shall let a, 6, and c represent
the lengths of the corresponding opposite sides.
18-2. THE LAW OF SINES
Law of Sines. Let ABC be any triangle lettered in the conven-
tional manner. Then the following relationship between the sides
and the sines of the angles may be written :
(18-1) . a = . b = c
sin A sin B sin C
This relationship is commonly called the law of sines.
B A
FIG. 18-1.
289
290 Solution of the General Triangle Sec. 1 8-2
Proof. We first note that all angles may be acute, as in Fig.
18-1 (a), or one angle, say B, may be obtuse, as in Fig. 18-1(6).
(The case where B = 90 entails no difficulties and will therefore
be omitted.) In each diagram, let h denote the altitude from
the vertex C to the side AB, Then, in either case, sin A = - and
h
sin B = - Dividing the first equation by the second, we have
a
sin A a a b
- _ j QJ - - --
sin B b sin A sin B
In a similar way, by drawing the altitude from the vertex A to
the side BC, we get
b c
sin B ~~ sin C
The equations thus obtained may be combined to give the law of
sines
"
sm A sm B sin C
Note. The law of sines is well adapted to the use of logarithms
because it involves only multiplications and divisions.
Since any pair of ratios in the law of sines involves two angles
and the sides opposite, it may be used in the solution of problems in
Cases I and II.
As noted above, the law of sines also applies in the special case
where ABC is a right triangle. In this case, one of the angles is
90, and the sine of that angle is 1.
18-3. SOLUTION OF CASE I BY THE LAW OF SINES: GIVEN ONE SIDE AND
TWO ANGLES
When one side and any two angles of a triangle are known, the
third angle can be found from the relation A + B + C = 180, and
each of the required sides is uniquely determined. These sides may
be found by the law of sines.
Example 18-1. In a triangle ABC, A = 3814', B = 6720', c = 329. Solve
the triangle.
Solution: The values of the given parts are indicated in Fig. 18-2. In this case,
c C = 180 - (3814' + 6720') = 7426'.
To find a, we use the relationship
a a 329
sin 3814 / sin 7426'
Therefore,
c-329 _ 329 sin 3814 /
FIG. 18-2. a "" sin 7426' '
Sec. 18-4
Solution of the General Triangle
291
and
log a = log 329 + log sin 3814' - log sin 7426'.
The indicated operations may be performed as follows:
log 329 = 2.5172
log sin 3S14' = 9.7916 - 10
log sin 7426' =
log a =
12.3088 - 10
9.9838 - 10
2.3250
a =211.
To determine 6, we use the relationship
329
Hence,
and
sin 6720' ~~ sin 7426 /
_ 329 sin 67 20'
"" sin 7426 / '
log b = log 329 + log sin 6720' - log sin 7426'.
The work follows:
log 329 = 2.5172
log sin 6720' = 9.9651 - 10
log sin 7426' =
log b =
12.4823 - 10
9.9838 - 10
2.4985
b = 315.
As shown in Fig. 18-1 (a),
c = b cos A + cos B.
This relationship may be used as a check. Thus,
c = 315 cos 3814' + 212 cos 6720'
= (315) (0.7855) + (212) (0.3854) = 329.1,
which agrees satisfactorily with the given value of c.
18-4. SOLUTION OF CASE II BY THE LAW OF SINES GIVEN TWO SIDES AND
THE ANGLE OPPOSITE ONE OF THEM
Case II is called the ambiguous case, because the data may be
such that two, one, or no triangles are determined. The number of
solutions when a, 6, and A are given is indicated by the accompany-
ing table.
TABLE OF POSSIBLE SOLUTIONS
(18-2)
a = b sin A
One right triangle
(18-3)
A ' APllfp
a < b sin A
No triangle
(18-4)
b sin A < a <b
Two triangles
(18-5)
a^b
One triangle
(18-6)
(18-7)
A obtuse
a^b
a>b
No triangle
One triangle
292
Solution of the General Triangle
Sec. 18-4
We shall use Fig. 18-3 to illustrate in turn the different possi-
bilities considered in the table. If the angle A and the sides a and b
are given, we first construct the angle A with the initial ray AX
and the terminal ray AR. Next, we lay off the distance AC = b
along the terminal side. Then, with C as the center and the length
of the side a as the radius, we describe an arc. We mark the point
or points in which this arc intersects the initial ray AX of the
angle A.
-X A
X
(<*)
\A
/B A
i) ()
FIG. 18-3.
Figure 18-3 (a) corresponds to (18-2) in the table. Since
BC = a = b sin A, this segment is the altitude of the triangle drawn
from the vertex C. Hence, the arc with radius a is tangent to the
initial side at B, and the triangle is a right triangle.
Figure 18-3(6) corresponds to (18-3) in the table. Since
a < b sin A, the side a is too short to intersect AX, and there is no
triangle.
In Fig. 18-3 (c), a < 6 and 6 sin A < a, as stated in (18-4) in the
table, and the arc will intersect AX in two points marked B and B'.
Therefore, two solutions exist. The angle B' in the triangle AB'C is
the supplement of the angle B in the triangle ABC.
Figure 18-3 (d) represents the case in which the side a is longer
than the side 6, as stated in (18-5) in the table. Hence, there is
only one point B in which the arc with radius a intersects the initial
ray AX of the angle A. There is only one solution.
Sec. 1 8-4 Solution of the General Triangle 293
In Fig. 18-3(e), the angle A is obtuse. Since a < b, as stated in
(18-6), the radius a is too short to intersect the initial ray AX,
and no triangle exists.
Finally, in Fig. 18-3 (/) , A is obtuse and a > 6, as stated in
(18-7). Here the arc can intersect the initial ray AX in only one
point. There is, in this case, only one triangle.
The following examples will illustrate some of the possibilities.
Example 18-2. In a triangle ABC, A = 3615', a = 9.8, b = 12.4. Solve the
triangle.
Solution: Draw Fig. 18-4 approximately to scale, showing the given parts. After
angle A and side 6 have been drawn, an arc is described with C as center and a as
radius. The arc intersects the side AX in two points, B\ and B 2 , and we apparently
have two possible triangles, ABiC and AB 2 C.
FIG. 18-4.
To find sin B, we use
sin B __ sin A
~T~ ~~~T~
Then 10 , . .
. ~ 12.4 sm 36
9.8
and
log sin B = log 12.4 + log sin 3615' - log 9.8
The logarithmic work follows:
log 12.4 = 1.0934
log sin 3615' = 9.7718 - 10
10.8652 - 10
log 9.8 = 0.9912
log sin B = 9.8740 - 10
B = 4826'.
There are two solutions, since 6 sin A < a < b. The left inequality follows from
the fact that log sin B = log sm < 0, and so sm , < 1. If we let BI = 4826',
294
Solution of the General Triangle
Sec. 18-4
then B 2 = 180 - 4826' = 13134' leads to another solution. Thus,
B l = 4826', Ci = 180 - (3615' + 4826') = 9519',
and
# 2 = ISl^', C 2 = 180 - (3615' + 13134') =
To find Ci, we use the law of sines again. Thus,
9.8 sin 9519 /
Similarly, we have
sin 3615'
9.8 sin 1211'
= 16.5.
= 3.5.
" 2 "" sin 3615'
As a partial check, the equation Ci = b cos A -f a cos B\ may be used. Thus,
12.4 cos 3615' + 9.8 cos 4826' = 12.4 (0.8064) + 9.8 (0.6635) = 16.5.
This result is the same as the value previously calculated. The same method may
be applied to check c 2 .
Example 18-3. Given A = 5630 , a = 13.0, b = 10.7, solve the triangle ABC.
Solution: Here we clearly have only one solution, since a > b. This can be seen
geometrically if we draw Fig. 18-5 approximately to scale and show the given parts.
Since a is greater than b, an arc with C as center intersects AX on opposite sides of
A. Obviously there is only one triangle, ABiC, containing the angle A.
To find BI, we use the relationship
sin BI sin 5630'
a - 13.0
10.7
13
whence
log sin
This gives
FIG. 18-5.
Then C = 180 - (5630' + 4320') = 8010'.
To find c, we use the law of sines and obtain
c 13
= log 10.7
-f log sin 5630' - log 13.
B l = 4320'.
sin 8010' sin 5630'
This gives c = 15.4.
Example 18-4. Given A = 6740', a = 16.0, b = 17.3, solve the triangle.
C Solution: The given parts are shown in Fig. 18-6. By
the law of sines, we have
sinff __ sin 6740 /
17.3 ~~ 16
a m 16.0 Therefore,
log sin B = log 17.3 -f log sin 6740' - log 16
= (1.2380) -f (9.9661 - 10) - (1.2041) = 0.
Hence, B = 90.
c B The solution may be completed by applying the theory
FIG. 18-6. of right triangles. Only one solution exists.
6740'
Sec. 18-4 Solution of the General Triangle 295
Example 18-5. Given A = 4723 ; , a = 230, 6 = 720, solve the triangle.
Solution: From Fig. 18-7, it appears that no triangle is possible. The following
work verifies this fact.
To find B t use the relationship
sin B __ sin A
btaining . *720sinV23',
Sm 230
The logarithmic work follows:
log 720 = 2.8573
log sin 4723' = 9.8668 - 10
12.7241 - 10
log 230 = 2.3617
log sin B = 10.3624 - 10 FlG * 18 ~ 7 ' -
= 0.3624.
Since log sin B > 0, sin B would have to be greater than 1, which is impossible.
Therefore, there is no solution.
EXERCISE 18-1
In each of the problems from 1 to 12, solve the given triangle by the law of sines.
1. a = 12.30, A = 3625', B = 4437'.
2. b = 12.18, A = 4733', B = 6751'.
3. c = 461.3, B = 6719', C = 2314'.
4. b = 0.6384, B = 3939', C = 8716'.
5. a = 6.714, A = 3753', C = 13636'.
6. c = 7832, A = Way, B = 4358'.
7. o = 21.23, c = 64.21, C = 6231 / .
8. 6 = 0.8146, c = 31.63, B = 1119'.
9. a = 987.4, b = 503.6, A = 5413 / .
10. a = 0.003862, c = 0.0008157, A = 2613'.
11. 6 = 1.386, c = 2.451, 5 = 83 19'.
12. 6 = 4.395, c = 9.806, C = 3746'.
13. A surveyor wishes to find the distance across a stream from point A to point B.
He finds that the distance from A to a point C on the same side of the stream is
687.4 feet, and angles BAG and BCA are 4953' and 816', respectively. Find
the distance AB.
14. A surveyor was running a line due west when he reached a swamp. From the
edge of the swamp he ran a line S 63 W for 2500 feet, and from this point he
ran a line N 2723' W. How far had he gone on this line when he reached his
original line produced? How far was it across the swamp?
15. A building 63.7 feet high stands on the top of a hill. From a point at the foot
of the hill the angles of elevation to the top and bottom of the building are
4216' and 3831', respectively. Find the height of the hill.
296
Solution of the General Triangle
Sec. 18-4
16. From a certain point on the ground the angle of elevation to the top of a building
is 46 17'. From a point on the ground 83 feet nearer the building the angle of
elevation is 6823'. Assuming that the ground is level, find the height of the
building.
17. One side and a diagonal of a parallelogram are 14.63 inches and 21.4 inches,
respectively. The angle between the diagonals and opposite the given side is
11623'. Find the length of the other diagonal.
18. It is necessary to measure the distance between two artillery pieces A and B.
The angle of depression from an observation point C to gun A is 2447'. Sound
travels at the rate of 1140 feet per second, and the sounds from guns A and B
reach C in 2.3 and 1.7 seconds, respectively. Find the distance AB, assuming
that points A, B, and C lie in the same plane.
19. A body is acted on by two forces, Fi = 2643 pounds and F 2 = 2341 pounds.
The resultant F 3 lies on a line making an angle of 4633' with Fi. Find F 3 and
the angle between the lines of action of Fi and F 2 . (The resultant of two
forces is their vector sum.)
20. A buoy, located at a point B, is 6 miles from a point A at one end of an island
and 10 miles from a point C at the other end of the island. If the angle BAG is
132 16', find the distance between the points A and C on the island.
18-5. THE LAW OF COSINES
Law of Cosines* Let ABC be any triangle. Then
(18-8) a 2 = b 2 + c 2 - 2bc cos A,
(18-9) b 2 = a 2 + c 2 - 2ac cos ,
(18-10) c 2 = a 2 + b 2 - 2ab cos C.
These relationships constitute the law of cosines.
FIG. 18-8.
Proof. We shall establish (18-8) by considering the case when A
is acute, as in Fig. 18-8 (a) and the case when A is obtuse as in
Fig. 18-8 (&). The case A = 90 involves no difficulty, and it will
therefore be omitted.
Let h denote the altitude from C to the side AB. Also, let x
denote the length AD. Hence, DB is c - x in Fig. 18-8 (a) and is
c + x in Fig. 18-8(6).
Sec. 1 8-6 Solution of fhe General Tr/ongfe 297
In Fig. 18-8 (a),
(c - xY + h 2 = a 2 ,
and
x 2 + h 2 = 6 2 .
Subtracting, we have c 2 2c# = a 2 6 2 , or
a 2 = 6 2 + c 2 - 2cz.
Since x = 6 cos A,
a 2 = 6 2 + c 2 - 26c cos A.
In Fig. 18-8(6),
(c + x) 2 + h 2 = a 2 ,
and
x 2 + h 2 = 6 2 .
Subtracting, we find that c 2 + 2cx = a 2 6 2 , or
a 2 = 6 2 + c 2 +
Since x = b cos A in this case,
a 2 = 6 2 + c 2 2bc cos A.
By drawing altitudes to the other sides and proceeding in a
similar manner, we obtain (18-9) and (18-10).
Note. For a simple algebraic proof of the law of cosines, see prob-
lem 22 in Exercise 18-4. The law of cosines applies equally well if
ABC is a right triangle. In this case, one of the formulas reduces
to the pythogorean theorem, since cos 90 = 0.
18-6. SOLUTION OF CASE III AND CASE IV BY THE LAW OF COSINES
Since the law of cosines is expressed by formulas involving addi-
tion and subtraction, it is not well adapted to logarithmic computa-
tion and its use is not recommended unless the given sides are
easily squared.
Example 18-6. Given b = 9.0, c = 13.0, A = 11510', solve the triangle ABC.
Solution: By the law of cosines,
a 2 = 52 _|_ C 2 _ 2bc cos A
= 9 2 + (13) 2 - 2(9) (13) cos 11510'
= 81 + 169 - 234( - cos 6450')
= 250 + 234 (0.4253) = 349.52.
Hence,
a = 18.7.
We employ the law of sines to find angle B. Thus,
sin _ sin 11510' _ sin 6450 /
9 "" 18.7 ~~ 18.7
5=25'50'.
Therefore, C = 180 - (11510' + 2550 / ) = 39.
298 Solution of the General Triangle Sec. 1 8-6
Example 18-7, Given a = 3, 6 = 5, c = 7. Find the angles.
Solution: From a 2 = 6 2 + c 2 26c cos 4, we have the formula
52 + C 2 _ a 2
cos A = - -T -
Therefore >
Hence, A = 21*47'.
Similarly, B = 3813' and C = 120.
We can check these by the equation A -f B + C = 180.
EXERCISE 18-2
In each of the problems from 1 to 6, solve the given triangle by the law of cosines.
l.o= 300, 6 = 250, C - 5840'. 2. a = 50, c = 240, 5 = 11050'.
3. 6 = 65, c = 310, A = 6710 / . 4. o = 130, c = 90, B = 10020'.
5. 6 = 50, c = 110, A = 150. 6. a = 1.63, 6 = 3.45, C = 2610'.
7. If a = 15, 6 = 12, c =5 20, find A.
8. If o = 25, b = 30, and c = 35, find B.
9. If a = 100, 6 = 300, c = 500, find C.
10. If a = 15, 6 = 12, and c = 20, find B.
11. If o = 16, 6 = 17, and c = 18, find A, B, C.
12. If o = 260, 6 = 322, c = 481, find A, B, C.
13. The distance between two points A and B cannot be measured directly.
Accordingly, a third point C is selected, and it is found that AC = 3000 feet,
BC = 4500 feet, and angle ACB = 4620'. Find the distance AB.
14. Two sides of a parallelogram are 125 feet and 200 feet, and the included angle is
11030'. Find the length of the longer diagonal of the parallelogram and also
the angle between that diagonal and a longer side of the parallelogram.
15. Two sides of a triangular plot of ground are 250 feet and 200 feet, and the
included angle is 6733'. Find the perimeter of the plot.
16. Two sides of a parallelogram are 700 feet and 420 feet, and one diagonal is
600 feet. Find the length of the other diagonal.
17. In a triangle ABC, a = 25, 6 = 27, and the median from A is 20. Find c, A, B, C.
18-7. THE LAW OF TANGENTS
Law of Tangents. Let ABC be any triangle. Then the following
relationships exist between two sides and the angles opposite them :
, tan i (A - B)
Ma in a ~ b - 2
(18-11) = = >
a + b tan (A + B)
Sec. 18-7 Solution of the General Triangle 299
(18-12)
tan (B - C)
b + c tan ^ (B + C)
(18-13) 5_IL = 2
C + a x 1 /^Y , X \
tan s (C + A)
These relationships constitute the law of tangents.
Proof. Let us denote the common ratio of the law of sines by r.
Thus, a = r sin A, 6 = r sin JS, c = r sin C. Then
a b _ r sin A r sin J? __ sin A sin B ^
a + 6 ~~ r sin A + r sin B ~" sin A + sin B
Substituting from (7-27) and (7-28) of Section 7-4, we have
. A . D 2 cos (A + B) sin (A - J5)
sm A sm g 2 y 2 '
Hence,
sin A + sin B ~~ . 1 , , , _. 1 , .
2 sin - (A + JS) cos - (A -
A z
tan (A -
A similar procedure may be followed to prove (18-12) and
(18-13).
We shall now use the law of tangents to solve a triangle in which
two sides and the included angle are given. Note that this law is
well adapted to logarithmic computation.
Example 18-8. Given 6 = 249, c = 372, A = 5622 ; , solve the triangle ABC.
Solution: To find C - B, we use the formula
Here c - 6 = 123, c + b = 621, C + B = 180 - A = 12338', (C + B) =
Therefore,
tan i (C - B) = g tan 6149',
and 1
log tan i (C - B) = log 123 + log tan 6149' - log 621.
300 Solution of f/ia Genera/ Triangle Sec. 1 8-7
The logarithmic work follows:
log 123 = 2.0899
log tan 6149' = 0.2710
12.3609 - 10
log 621 = 2.7931
log tan i (C - B) = 9.5678 - 10
i
1 (C - B) = 2017'.
Hence, - 1
C=i(C+)+i(C-)=826',
and
B = ~ (C + ) - ~ (C - B) = 4132'.
Cfodk:
4 4. 5 + C = 5622' + 826 / + 4132 7 = 180.
To find a, we use the law of sines. Thus,
249 sin 6622'
sin 4132'
Q1Q
=313 '
18-8. THE HALF-ANGLE FORMULAS
The following relationships are very convenient for the loga-
rithmic solution of Case IV, where the three sides a, 6, and c are
known :
- 6)
2 r (* c)
In these formulas, 5 = - (a + b + c).
2
Proof. From (7-16) in Section 7-3, we have
A l-cosA
2 ~ 1 + cos A
Also, from the law of cosines,
6 2 + c 2 - a 2
cos A = - 25^ --
Therefore,
_i & 2 + c 2 - a 2 _ a 2 - (6 - c) 2 _ (a + b - c) (a - b + c)
-l - ~
2bc ~ 2bc
and
Sec. 18-8 Solution of the General Triangle 301
If we let a + b + c = 2s, then
a + b - c = a + 6 + c - 2c = 2(a - c),
a - 6 + c = a + 6 + c - 26 = 2(s - 6),
6 + c - a = a + 6 + c - 2a = 2(s - a).
Therefore,
2 .A _ (a + & - c) (a - 6 + c) _ (g - c) ( - 6)
2 (6 + c + a) (b + c - a) " t( - a)
and
(18-14)
v ' 2 r s(s a)
We can derive (18-15) and (18-16) in a similar manner.
Example 18-9. Given: a = 379, b = 227, c - 416, find the angles of the triangle.
Solution: Here 2s=a+b+c = 1022. Then
8 =511,
s - a = 132,
s - 6 = 284,
s - c = 95.
Hence,
. A A /(284) (95)
-
The calculations by logarithms follow:
log 284 = 2.4533 log 511 = 2.7084
log 95 = 1.9777 log 132 = 2.1206
log (284) (95) = 24.4310 - 20 4.8290
log (511) (132) = 4.8290
19.6020 - 20
log tan y = 9.8010 - 10
. ., , A = 6438 ; .
Similarly, .
B _ ^7
2--r
(132) (95)
(511) (284)'
C _ y |/(132)(284)
w " 2 "" V (511) (95)
Hence, we find that B = 3246' and C = 8236 / .
A + 5 + C = 6438' + 3246' + 8236' = 180.
EXERCISE 18-3
In each of the problems from 1 to 10, solve the given triangle by the law of
tangents if an angle is given, or by the half-angle formulas if three sides are given
1. a = 50, b = 60, C = 60. 2. 6 = 17.1, c = 22.3, A = 2116 / .
3. a = 230, c = 106, B = 9510 / . 4. 6 = 79.3, c = 113, A =
302 Solution of the General Triangle Sec. 1 8-8
5. 6 = 41.82, c = 75.89, A = 7849'. 6. a = 0.1028, 6 = 0.8726, C = 14S13'.
7. a = 625, 6 = 725, c = 825. 8. a = 60.65, 6 = 38.64, c = 23.57.
9. a = 67450, 6 = 84380, c = 98630. 10. a = 0.1146, 6 = 0.3184, c = 0.6379.
11. The diagonals of a parallelogram are 6 inches and 10 inches, and they intersect
at an angle of 63. Find the sides of the parallelogram.
12. Points A and B are separated by an obstacle. In order to find the distance
between them, a third point C is selected and it is found that AC = 126 rods
and BC = 185 rods. The angle subtended at C by AB is 9614'. Find AB.
13. Two circles whose radii are 14 and 17 inches respectively intersect. The angle
between the tangents to the circles at either point of intersection is 3846'.
Find the distance between the centers of the circles.
14. The sides of a parallelogram are 13.4 inches and 18.5 inches, and one diagonal
is 15.6 inches. Find the angles and the other diagonal of the parallelogram.
15. Three circles whose radii are 10, 11, and 12 inches, respectively, are tangent to
each other externally. Find the angles of the triangle formed by joining their
centers.
16. The sides of a triangular field are in the proportion 4:5:6. The area of the field is
18 acres. If there are 160 square rods in an acre, find the length of each side
of the field in rods.
17. In triangle ABC, prove the following:
a -6 sin ^-*)
I n
cos-C
a +b -s
C . 1
sm-C
These formulas are called Mollweide's equations. They may be used in checking
the solution of a triangle.
18-9. AREA OF A TRIANGLE
We can readily see that the area K of the triangle in Fig. 18-1 is
(18-17) K = \ch = JcbsinA.
& <u
In either triangle, h = 6 sin A. In like manner, we obtain
(18-18) K = lac sin B,
Zi
(18-19) K = ^ab sin C.
By substituting c sm - for 6 from the law of sines, we may
sm C
transform (18-17) to obtain
n o(\\ K - c * sin A sin B
(18 ~ 20) K ~ 2sinC
Sec. 1 8-9 Solution of the General Triangle 303
By cyclic interchanges of letters, we obtain
(18-21) K = a2 sin B sin C
(18-22) K =
2 sin A
b 2 sin A sin C
2smB
To derive a formula for finding the area of a triangle when its
three sides are given, we first transform (18-17) in the following
manner : '
K = ybc sin A = ^bc sin (2 y J
1,/ . A A
= rpc (2 sm y cos 2
, . A A
= oc sm -jr- cos -^-
Zi
But, from (7-14) and (7-15) in Section 7-3,
A /I cos A , A ^ /I
= v 2 and cos 2" = y
2
Using the values from Section 18-8 for 1 cos A and 1 -f cos A, we
have
. A A /(s c) (s 6) , A . /s(s a)
sm - A/ ^ '- and cos -^ - \/ --r -
2 V be 2 V be
Consequently, the formula for area in terms of the sides is
= \/ s ( s a) (s t) (s c).
The following examples illustrate the use of the area formulas.
Example 18-10. Find the area of the triangle in Example 18-8, in which b ='249,
c = 372, A = 5622'.
Solution: Since two sides and the included angle are given, (18-17) may be used*
Thus,
K = ^bc sin A = ^(249) (372) sin 5622 / .
2 Z
The logarithmic work follows :
log 0.5 = 9.6990 - 10
log 249 = 2.3962
log 372 = 2.5705
log sin 5622' = 9.9205 - 10
log K = 24.5862 - 20
K = 38,600.
304 Solution of the General Triangle Sec. 1 8-9
Example 18-11. Find the area of the triangle in Example 18-1, in which A
= 3814', B = 6720', c = 329.
Solution: Since two angles and a side are given, (18-20) may be used. In this case,
^ _ c 2 sin A sin B _ (329)* sin 3814' sin 6720' _ Q0 inn
* - 2 sinC ~ 2 sin 7426' "" * 2 ' 100 '
Example 18-12. Find the area of the triangle in Example 18-9, in which a = 379,
6 = 227, c = 416.
Solution: In this solution, (18-21) is used. We have
K = Vs(s - a) (s - 6) ( - c) = V(511) (132) (284) (95) = 42,700.
EXERCISE 18-4
In each of the problems from 1 to 8, find the area of the given triangle.
1. a = 12.30, A = 3625', B = 4437'. 2. c = 461.3, B = 6719', C = 2314 / .
3. a = 987.4, b = 503.6, A = 5413'. 4. 6 = 4.395, c = 9.806, C = 3746'.
5. 6 = 65, c = 310, A = 6710'. 6. a = 300, 6 = 250, C = 5840'.
7. a = 15, b = 12, c = 20. 8. a = 100, 6 = 300, c = 500.
9. In triangle ABC, let r be the radius of the inscribed circle. Prove that K = rs
j A u r j.1. x /4 /
and, therefore, that r = /I/
r
($ a) ( 6) (S - C)
- ~ - - -
10. Find the radius of the circle inscribed in the triangle whose sides are 48.92
feet, 63.86 feet, and 72.31 feet.
11. A cylindrical tank is to be built on a triangular lot having sides whose lengths
are 200 feet, 186 feet, and 176 feet. Find the radius of the largest such tank
which can be built on the lot.
12. In triangle ARC, let R be the radius of the circumscribed circle. Show that
2B= a = b = *
sin A sin B sin C
13. In triangle ABC, show that R = -r^ > where R is the radius of the circum-
scribed circle and K is the area of the triangle.
14. The sides of a triangle are 23, 29, and 46 feet. Find the areas of the triangle
and the inscribed and circumscribed circles.
15. The sides of a triangular plot of grass are 42 feet, 65 feet, and 87 feet. Find the
minimum radius of action of an automatic lawn sprinkler which will water all
parts of the plot from the same point.
16. An arc of a circle of radius r subtends a central angle 0. Show that the area
bounded by this arc and its chord is jr r 2 (6 - sin 8).
t
17. Find the area of the largest pentagon which can be cut from a circular piece of
metal 4 feet in radius. How much metal is wasted?
18. In triangle ABC, prove that the median from any vertex to the side opposite
divides the angle at that vertex into two parts whose sines are proportional to
the lengths of the parts into which the side opposite is divided by the median.
Sec. 1 8-9 Solution of the General Triangle 305
19. In triangle ABC, prove that
cos A cos B cos C _ a 2 -f fr 2 -f c 2 1
a + 6 + c "" 2a6c
20. In triangle ABC, prove that
a + b + c = (6 + c) cos A + (c + a) cos B + (a + b) cos C.
21. In triangle ABC, prove that
a 2 -f 6 2 + c 2 = a 2 (cos 2 C + sin 2 B) + 6 2 (cos 2 4 -f sin 2 C) + c 2 (cos 2 B + sin 2 A).
22. In triangle ABC, show that
a = b cos (7 -f c cos 5,
6 = c cos A -f a cos C,
c = a cos B + 6 cos A.
Multiply the first equation by a, the second by 6, and the third by c, to give a
second proof of the law of cosines; that is, prove (18-8) by showing that
6 2 + c 2 - a 2 = 2bc cos A. Similarly prove (18-9) and (18-10).
23. Consider any triangle ABC. If a > b, prove that A > B. If A > B, prove that
a > b.
24. In triangles ABC and A'B'C', let A and A', B and B', C and C" be pairs of
corresponding vertices, and let the corresponding sides be a and a', b and b',
c and c'. If a = a ; , 6 = 6', and C > C', prove that c > c'. If a = a', 6 = &',
and c > c', prove that C > C'.
Appendix
A Tables
Appendix A
30*
TABLE I
FOUR-PLACE VALUES OF FUNCTIONS OF NUMBERS
t
sin t
cos t
tant
cot t
sec t
CSC t
00
.0000
1.0000
.0000
1.000
W/
.01
'.0100
1.0000
.0100
99.997
1.000
100.00
.02
.0200
.9998
.0200
49.993
1.000
50.00
.03
.0300
.9996
.0300
33.323
1.000
33.34
.04
.0400
.9992
.0400
24.987
1.001
25.01
.05
.0500
.9988
.0500
19.983
1.001
20.01
.06
.0600
.9982
.0601
16.647
1.002
16.68
.07
.0699
.9976
.0701
14.262
1.002
14.30
.08
.0799
.9968
.0802
12.473
1.003
12.51
.09
.0899
.9960
.0902
11.081
1.004
11.13
.10
.0998
.9950
.1003
9.967
1.005
10.02
.11
.1098
.9940
.1104
9.054
1.006
9.109
.12
.1197
.9928
.1206
8.293
1.007
8.353
.13
.1296
.9916
.1307
7.649
1.009
7.714
.14
.1395
.9902
.1409
7.096
1.010
7.166
.15
.1494
.9888
.1511
6.617
1.011
6.692
.16
.1593
.9872
.1614
6.197
1.013
6.277
.17
.1692
.9856
.1717
5.826
1.015
5.911
.18
.1790
.9838
.1820
5.495
1.016
5.586
.19
.1889
.9820
.1923
5.200
1.018
5.295
.20
.1987
.9801
.2027
4.933
1.020
5.033
.21
.2085
.9780
.2131
4.692
1.022
4.797
.22
.2182
.9759
.2236
4.472
1.025
4.582
.23
.2280
.9737
.2341
4.271
1.027
4.386
.24
.2377
.9713
.2447
4.086
1.030
4.207
.25
.2474
.9689
.2553
3.916
1.032
4.042
.26
.2571
.9664
.2660
3.759
1.035
3.890
.27
.2667
.9638
.2768
3.613
1.038
3.749
.28
.2764
.9611
.2876
3.478
1.041
3.619
.29
.2860
.9582
.2984
3.351
1.044
3.497
.30
.2955
.9553
.3093
3.233
1.047
3.384
.31
.3051
.9523
.3203
3.122
1.050
3.278
.32
.3146
.9492
.3314
3.018
1.053
3.179
.33
.3240
.9460
.3425
2.920
1.057
3.086
.34
.3335
.9428
.3537
2.827
1.061
2.999
.35
.3429
.9394
.3650
2.740
1.065
2.916
.36
.3523
.9359
.3764
2.657
1.068
2.839
.37
.3616
.9323
.3879
2.578
1.073
2.765
.38
.3709
.9287
.3994
2.504'
1.077
2.696
.39
.3802
.9249
.4111
2.433
1.081
2.630
t
sin t
cos t
tan t
cot t
sec t
CSC t
310
Appendix A
TABLE I (continued)
t
sin t
cos I
tan t
cot t
sec t
CSC t
.40
.3894
.9211
.4228
2.365
1.086
2.568
.41
.3986
.9171
.4346
2.301
1.090
2.509
.42
.4078
.9131
.4466
2.239
1.095
2.452
.43
.4169
.9090
.4586
2.180
1.100
2.399
.44
.4259
.9048
.4708
2.124
1.105
2.348
.45
.4350
.9004
.4831
2.070
1.111
2.299
.46
.4439
.8961
.4954
2.018
1.116
2.253
.47
.4529
.8916
.5080
1.969
1.122
2.208
.48
.4618
.8870
.5206
1.921
1.127
2.166
.49
.4706
.8823
.5334
1.875
1.133
2.125
.50
.4794
.8776
.5463
1.830
1.139
2.086
.51
.4882
.8727
.5594
1.788
1.146
2.048
.52
.4969
.8678
.5726
1.747
1.152
2.013
.53
.5055
.8628
.5859
1.707
1.159
1.978
.54
.5141
.8577
.5994
1.668
1.166
1.945
.55
.5227
.8525
.6131
1.631
1.173
1.913
.56
.5312
.8473
.6269
1.595
1.180
1.883
.57
.5396
.8419
.6410
1.560
1.188
1.853
.58
.5480
.8365
.6552
1.526
1.196
1.825
.59
.5564
.8309
.6696
1.494
1.203
1.797
.60
.5646
.8253
.6841
1.462
1.212
1.771
.61
.5729
.8196
.6989
1.431
1.220
1.746
.62
.5810
.8139
.7139
1.401
1.229
1.721
.63
.5891
.8080
.7291
1.372
1.238
1.697
.64
.5972
.8021
.7445
1.343
1.247
1.674
.65
.6052
.7961
.7602
1.315
1.256
1.652
.66
.6131
.7900
.7761
1.288
1.266
1.631
.67
.6210
.7838
.7923
1.262
1.276
1.610
.68
.6288
.7776
.8087
1.237
1.286
1.590
.69
.6365
.7712
.8253
1.212
1.297
1.571
.70
.6442
.7648
.8423
1.187
1.307
1.552
.71
.6518
.7584
.8595
1.163
1.319
1.534
.72
.6594
.7518
.8771
1.140
1.330
1.517
.73
.6669
.7452
.8949
1.117
1.342
1.500
.74
.6743
.7385
.9131
1.095
1.354
1.483
.75
.6816
.7317
.9316
1.073
1.367
1.467
.76
.6889
.7248
.9505
1.052
1.380
1.452
.77
.6961
.7179
.9697
1.031
1.393
1.437
.78
.7033
.7109
.9893
1.011
1.407
1.422
.79
.7104
.7038
1.009
.9908
1.421
1.408
t
sin t
cos t
tan t
cot t
sec t
CSC t
Appendix A
311
TABLE I (continued)
t
sin t
cos t
tan t
cot t
sec t
CSC t
.80
.7174
.6967
1.030
.9712
1.435
1.394
.81
.7243
.6895
1.050
.9520
1.450
1.381
.82
.7311
.6822
1.072
.9331
1.466
1.368
.83
.7379
.6749
1.093
.9146
1.482
1.355
.84
.7446
.6675
1.116
.8964
1.498
1.343
.85
.7513
.6600
1.138
.8785
1.515
1.331
.86
.7578
.6524
1.162
.8609
1.533
1.320
.87
.7643
.6448
1.185
.8437
1.551
1.308
.88
.7707
.6372
1.210
.8267
1.569
1.297
.89
.7771
.6294
1.235
.8100
1.589
1.287
.90
.7833
.6216
1.260
.7936
1.609
1.277
.91
.7895
.6137
1.286
.7774
1.629
1.267
.92
.7956
.6058
1.313
.7615
1.651
1.257
.93
.8016
.5978
1.341
.7458
1.673
1.247
.94
.8076
.5898
1.369
.7303
1.696
1.238
.95
.8134
.5817
1.398
.7151
1.719
1.229
.96
.8192
.5735
1.428
.7001
1.744
1.221
.97
.8249
.5653
1.459
.6853
1.769
1.212
.98
.8305
.5570
1.491
.6707
1.795
1.204
.99
.8360
.5487
1.524
.6563
1.823
1.196
1.00
.8415
.5403
1.557
.6421
1.851
1.188
1.01
.8468
.5319
1.592
.6281
1.880
1.181
1.02
.8521
.5234
1.628
.6142
1.911
1.174
1.03
.8573
.5148
1.665
.6005
1.942
1.166
1.04
.8624
.5062
1.704
.5870
1.975
1.160
1.05
.8674
.4976
1.743
.5736
2.010
1.153
1.06
.8724
.4889
1.784
.5604
2.046
1.146
1.07
.8772
.4801
1.827
.5473
2.083
1.140
1.08
.8820
.4713
1.871
.5344
2.122
1.134
1.09
.8866
.4625
1.917
.5216
2.162
1.128
1.10
.8912
.4536
1.965
.5090
2.205
1.122
1.11
.8957
.4447
2.014
.4964
2.249
1.116
1.12
.9001
.4357
2.066
.4840
2.295
1.111
1.13
.9044
.4267
2.120
.4718
2.344
1.106
1.14
.9086
.4176
2.176
.4596
2.395
1.101
1.15
.9128
.4085
2.234
.4475
2.448
1.096
1.16
.9168
.3993
2.296
.4356
2.504
1.091
1.17
.9208
.3902
2.360
.4237
2.563
1.086
1.18
.9246
.3809
2.427
.4120
2.625
1.082
1.19
.9284
.3717
2.498
.4003
2.691
1.077
t
sin t
cos t
tan t
cot t
sec t
CSC t
Appendix A
TABLE I (continued)
t
sin t
cos t
tan t
cot t
sec t
CSC t
1.20
.9320
.3624
2.572
.3888
2.760
1.073
1.21
.9356
.3530
2.650
.3773
2.833
1.069
1.22
.9391
.3436
2.733
.3659
2.910
1.065
1.23
.9425
.3342
2.820
.3546
2.992
1.061
1.24
.9458
.3248
2.912
.3434
3.079
1.057
1.25
.9490
.3153
3.010
.3323
3.171
1.054
1.26
.9521
.3058
3.113
.3212
3.270
1.050
1.27
.9551
.2963
3.224
.3102
3.375
1.047
1.28
.9580
.2867
2.341
.2993
3.488
1.044
1.29
.9608
.2771
3.467
.2884
3.609
1.041
1.30
.9636
.2675
3.602
.2776
3.738
1.038
1.31
.9662
.2579
3.747
.2669
3.878
1.035
1.32
.9687
.2482
3.903
.2562
4.029
1.032
1.33
.9711
.2385
4.072
.2456
4.193
1.030
1.34
.9735
.2288
4.256
.2350
4.372
1.027
1.35
.9757
.2190
4.455
.2245
4.566
1.025
1.36
.9779
.2092
4.673
.2140
4.779
1.023
1.37
.9799
.1994
4.913
.2035
5.014
1.021
1.38
.9819
.1896
5.177
.1931
5.273
1.018
1.39
.9837
.1798
5.471
.1828
5.561
1.017
1.40
.9854
.1700
5.798
.1725
5.883
1.015
1.41
.9871
.1601
6.165
.1622
6.246
1.013
1.42
.9887
.1502
6.581
.1519
6.657
1.011
1.43
.9901
.1403
7.055
.1417
7.126
1.010
1.44
.9915
.1304
7.602
.1315
7.667
1.009
1.45
.9927
.1205
8.238
.1214
8.299
1.007
1.46
.9939
.1106
8.989
.1113
9.044
1.006
1.47
.9949
.1006
9.887
.1011
9.938
1.005
1.48
.9959
.0907
10.983
.0910
11.D29
1.004
1.49
.9967
.0807
12.350
.0810
12.390
1.003
1.50
.9975
.0707
14.101
.0709
14.137
1.003
1.51
.9982
.0608
16.428
.0609
16.458
1.002
1.52
.9987
.0508
19.670
.0508
19.695
1.001
1.53
.9992
.0408
24.498
.0408
24.519
1.001
1.54
.9995
.0308
32.461
.0308
32.476
1.000
1.55
.9998
.0208
48.078
.0208
48.089
1.000
1.56
.9999
.0108
92.620
.0108
92.626
1.000
1.57
1.0000
.0008
1255.8
.0008
1255.8
1.000
1.58
1.0000
-.0092
i -108.65
-.0092
-108.65
1.000
1.59
.9998
-.0192
-52.067
-.0192
-52.08
1.000
1.60
. .9996
-.0292
-34.233
-.0292
-34.25
1.000
t
sin t
cos*
tan t
cot t
sec t
CSC t
Appendix A
313
TABLE II
FOUR-PLACE VALUES OP FUNCTIONS
*
Sin
Cos
Tan
Cot
Sec
Csc
AO AA/
noon
1 ftftn
noon
1 noo
QAO AA/
V VV
vl/UU
l.UUU
UUVU
l.UUU
v Uw
10'
20'
30'
40'
050'
029
058
.0087
116
145
000
000
1.000
.9999
999
029
058
.0087
116
145
343.8
171.9
114.6
85.94
68.75
000
000
1.000
000
000
343.8
171.9
114.6
85.95
68.76
89 50'
40'
30'
20'
10'
100'
.0175
.9998
.0175
57.29
1.000
57.30
89 00'
10'
20'
30'
40'
150'
204
233
.0262
291
320
998
997
.9997
996
995
204
233
.0262
291
320
49.10
42.96
38.19
34.37
31.24
000
000
1.000
000
001
49.11
42.98
38.20
34.38
31.26
88 50'
40'
30'
20'
10'
2 00'
.0349
.9994
.0349
28.64
1.001
28.65
88 00'
10'
20'
30'
40'
2 50'
378
407
.0436
465
494
993
992
.9990
989
988
378
407
.0437
466
495
26.43
24.54
22.90
21.47
20.21
001
001
1.001
001
001
26.45
24.56
22.93
21.49
20.23
87 50'
40'
30'
20'
10'
3 00'
.0523
.9986
.0524
19.08
1.001
19.11
87 00'
10'
20'
30'
40'
3 50'
552
581
.0610
640
669
985
983
.9981
980
978
553
582
.0612
641
670
18.07
17.17
16.35
15.60
14.92
002
002
1.002
002
002
18.10
17.20
16.38
15.64
14.96
86 50'
40'
30'
20'
10'
4 00'
.0698
.9976
.0699
14.30
1.002
14.34
86 00'
10'
20'
30'
40'
4 50'
727
756
.0785
814
843
974
971
.9969
967
964
729
758
.0787
816
846
13.73
13.20
12.71
12.25
11.83
003
003
1.003
003
004
13.76
13.23
12.75
12.29
11.87
85*50'
40'
30'
20'
10'
5 oo'
.0872
.9962
.0875
11.43
1.004
11.47
86 00'
10'
20'
30'
40'
5 50'
901
929
.0958
.0987
.1016
959
957
.9954
951
948
904
934
.0963
.0992
.1022
11.06
10.71
10.39
10.08
9.788
004
004
1.005
005
005
11.10
10.76
10.43
10.13
9.839
8450 /
40'
30'
20?
W
6 00>
.1045
.9945
.1051
9.514
1.006
9.567
84 W
Cot
Sin
Cot
Tan
Csc
Sec
+
314
Appendix A
TABLE II (continued)
^
Sin
Cos
Tan
Cot
Sec
Csc
6 00'
.1045
.9945
.1051
9.514
1.006
9.567
84 00'
Itf
20'
30'
40
6 50'
074
103
.1132
161
190
942
939
.9936
932
929
080
110
.1139
169
198
255
9.010
8.777
556
345
006
006
1.006
007
007
309
9.065
8.834
614
405
83 50'
40'
30'
20'
10'
7 W
.1219
.9925
.1228
8.144
1.008
8.206
83 00'
10'
20'
30'
40'
7 50'
248
276
.1305
334
363
922
918
.9914
911
907
257
287
.1317
346
376
7.953
770
7.596
429
269
008
008
1.009
009
009
8.016
7.834
7.661
496
337
82 50'
40'
30'
20'
10'
800 /
.1392
.9903
.1405
7.115
1.010
7.185
82 00'
10'
20'
30'
40'
8 50'
421
449
.1478
507
536
899
894
.9890
886
881
435
465
.1495
524
554
6.968
827
6.691
561
435
010
Oil
1.011
012
012
7.040
6.900
6.765
636
512
81 50'
40'
30'
20'
10'
9 00'
.1564
.9877
.1584
6.314
1.012
6.392
81 00'
10'
20'
30'
40'
9 50'
593
622
.1650
679
708
872
868
.9863
858
853
614
644
.1673
703
733
197
6.084
5.976
871
769
013
013
1.014
014
015
277
166
6.059
5.955
855
80 50'
40'
30'
20'
10'
10 00'
.1736
.9848
.1763
5.671
1.015
5.759
80 00'
10'
20'
30'
40'
10 50'
765
794
.1822
851
880
843
838
.9833
827
822
793
823
.1853
883
914
576
485
5.396
309
226
016
016
1.017
018
018
665
575
5.487
403
320
79 50'
40'
30'
20'
10'
11 W
.1908
.9816
.1944
5.145
1.019
5.241
79 00'
10'
20'
30'
40'
11* 50'
937
965
.1994
.2022
051
811
805
.9799
793
787
.1974
.2004
.2035
065
095
5.066
4.989
4.915
843
773
019
020
1.020
021
022
164
089
5.016
4.945
876
78 50'
40'
30'
20'
10'
12 00'
.2079
.9781
.2126
4.705
1.022
4.810
78 00'
Cos
Sin
Cot
Tan
Csc
Sec
<
Appendix A
315
TABLE II (continued)
*.
Sin
Cos
Tan
Cot
Sec
Csc
12 00'
.2079
.9781
.2126
4.705
1.022
4.810
78 00'
10'
20'
30'
40'
12 50'
108
136
.2164
193
221
775
769
.9763
757
750
156
186
.2217
247
278
638
574
4.511
449
390
023
024
1.024
025
026
745
682
4.620
560
502
77 50'
40'
30'
20'
10'
13 00'
.2250
.9744
.2309
4.331
1.026
4.445
77 00'
10'
20'
30'
40'
13 50'
278
306
.2334
363
391
737
730
.9724
717
710
339
370
.2401
432
462
275
219
4.165
113
061
027
028
1.028
029
030
390
336
4.284
232
182
76 50'
40'
30'
20'
10'
14 00'
.2419
.9703
.2493
4.011
1.031
4.134
76 00'
10'
20'
30'
40'
14 50'
447
476
.2504
532
560
696
689
.9681
674
667
524
555
.2586
617
648
3.962
914
3.867
821
776
031
032
1.033
034
034
086
4.039
3.994
950
906
75 50'
40'
30'
2<X
10'
15 00'
.2588
.9659
.2679
3.732
1.035
3.864
75 00'
10'
20'
30'
40'
15 50'
616
644
.2672
700
728
652
644
.9636
628
621
711
742
.2773
805
836
689
647
3.606
566
526
036
037
1.038
039
039
822
782
3.742
703
665
74 SO 7
40'
30'
20'
10'
16 00'
.2756
.9613
.2867
3.487
1.040
3.628
74 00'
10'
20'
30'
40'
16 50'
784
812
.2840
868
896
605
596
.9588
580
572
899
931
.2962
.2994
.3026
450
412
3.376
340
305
041
042
1.043
044
045
592
556
3.521
487
453
73 50'
40'
30'
20'
10'
17 00'
.2924
.9563
.3057
3.271
1.046
3.420
730 00'
10'
20'
30'
40'
17 50'
952
.2979
.3007
035
062
555
546
.9537
528
520
089
121
.3153
185
217
237
204
3.172
140
108
047
048
1.049
049
050
388
356
3.326
295
265
72 50'
40'
30'
20'
10'
18 00'
.3090
.9511
.3249
3.078
1.051
3.236
720<K
Cos
Sin
Cot
Tan
Csc
Sec
<
316
Appendix A
TABLE II (continued)
^
Sin
Cos
Tan
Cot
Sec
Csc
18 00'
.3090
.9511
.3249
3.078
1.051
3.236
72 00'
10'
20'
30'
. 40'
18 50'
118
145
.3173
201
228
502
492
.9483
474
465
281
314
.3346
378
411
047
3.018
2.989
960
932
052
053
1.054
056
057
207
179
3.152
124
098
71 50'
40'
30'
20'
10'
19 00'
.3256
.9455
.3443
2.904
1.058
3.072
71 00'
10>
20'
30'
40'
19 50'
20 00'
283
311
.3338
365
393
.3420
446
436
.9426
417
407
.9397
476
508
.3541
574
607
.3640
877
850
2.824
798
773
2.747
059
060
1.061
062
063
1.064
046
3.021
2.996
971
947
2.924
70 50'
40'
30'
20'
10'
70 00'
10'
20'
30'
40'
20 50'
448
475
.3502
529
557
387
377
.9367
356
346
673
706
.3739
772
805
723
699
2.675
651
628
065
066
1.068
069
070
901
878
2.855
833
812
69 50'
40'
30'
20'
10'
21 00'
.3584
.9336
.3839
2.605
1.071
2.790
69 00'
10'
20'
30'
40'
21 50'
22 00'
611
638
.3665
692
719
.3746
325
315
.9304
293
283
.9272
872
906
.3939
.3973
.4006
.4040
583
560
2.539
517
496
2.475
072
074
1.075
076
077
1.079
769
749
2.729
709
689
2.669
68 50'
40'
30'
20'
10'
68 00'
10'
20'
30'
40'
22 50'
773
800
.3827
854
881
261
250
.9239
228
216
074
108
.4142
176
210
455
434
2.414
394
375
080
081
1.082
084
085
650
632
2.613
595
577
67 50'
40'
30'
20'
10'
23 00'
.3907
,9205
.4245
2.356
1.086
2.559
67 00'
10'
20'
30'
40'
23 50'
934
961
.3987
.4014
041
194
182
.9171
159
147
279
314
.4348
383
417
337
318
2.300
282
264
088
089
1.090
092
093
542
525
2.508
491
475
66 50'
40'
30'
20'
10'
24 00'
.4067
Cos
.9135
Sin
.4452
Cot
2.246
Tan
1.095
Csc
2.459
Sec
66 00'
Appendix A
317
TABLE II (continued)
>.
Sin
Cos
Tan
Cot
Sec
Csc
24 00'
.4067
.9135
.4452
2.246
1.095
2.459
66 00'
10'
20'
30'
40'
24 50'
094
120
.4147
173
200
124
112
.9100
088
075
487
522
.4557
592
628
229
211
2.194
177
161
096
097
1.099
100
102
443
427
2.411
396
381
65 50'
40'
30'
20'
10'
25 00'
.4226
.9063
.4663
2.145
1.103
2.366.
65 00'
10'
20'
30'
40'
25 50'
253
279
.4305
331
358
051
038
,9026
013
.9001
699
734
.4770
806
841
128
112
2.097
081
066
105
106
1.108
109
111
352
337
2.323
309
295
64 50'
40'
30'
20'
10'
26 00'
.4384
.8988
.4877
2.050
1.113
2.281
64 00'
10'
20'
30'
40'
26 50'
410
436
.4462
488
514
975
962
.8949
936
923
913
950
.4986
.5022
059
035
020
2.006
1.991
977
114
116
1.117
119
121
268
254
2.241
228
215
63 50'
40'
30'
20'
10'
27 00'
.4540
.8910
.5095
1.963
1.122
2.203
63 00'
10'
20'
30'
40'
27 50'
566
592
.4617
643
669
897
884
.8870
857
843
132
169
.5206
243
280
949
935
1.921
907
894
124
126
1.127
129
131
190
178
2.166
154
142
62 50'
40'
30'
20'
10'
28 00'
.4695
.8829
.5317
1.881
1.133
2.130
62 00'
10'
20'
30'
40'
28 50'
720
746
.4772
797
823
816
802
.8788
774
760
354
392
.5430
467
505
868
855
1.842
829
816
134
136
1.138
140
142
118
107
2.096
085
074
61 50'
40'
30'
20'
10'
29 00'
.4848
.8746
.5543
1.804
1.143
2.063
61 00'
10'
20'
30'
40'
29 50'
874
899
.4924
950
.4975
732
718
.8704
689
675
581
619
.5658
696
735
792
780
1.767
756
744
145
147
1.149
151
153
052
041
2.031
020
010
60 50'
40'
30'
20'
10'
30 00'
.5000
.8660
.5774
1.732
1.155
2.000
60 <W
Cos
Sin
Cot
Tan
Csc
Sec
318
Appendix A
TABLE II (continued)
^
Sin
Cos
Tan
Cot
Sec
Csc
30 00'
.5000
.8660
.5774
1.732
1.155
2.000
60 00'
10'
20'
30'
40'
30 50'
025
050
.5075
100
125
646
631
.8616
601
587
812
851
.5890
930
.5969
720
709
1.698
686
675
157
159
1.161
163
165
1.990
980
1.970
961
951
59 50'
40'
30'
20'
10'
31 00'
.5150
.8572
.6009
1.664
1.167
1.942
59 00'
10'
20'
30'
40'
31 50'
175
200
.5225
250
275
557
542
.8526
511
496
048
088
.6128
168
208
653
643
1.632
621
611
169
171
1.173
175
177
932
923
1.914
905
896
58 50'
40'
30'
20'
10'
32 00'
.5299
.8480
.6249
1.600
1.179
1.887
58 00'
10'
20'
30'
40'
32 50'
324
348
.5373
398
422
465
450
.8434
418
403
289
330
.6371
412
453
590
580
1.570
560
550
181
184
1.186
188
190
878
870
1.861
853
844
57 50'
40'
30'
20'
10'
33 00'
.5446
.8387
.6494
1.540
1.192
1.836
67 00'
10'
20'
30'
40'
33 50'
471
495
.5519
544
568
371
355
.8339
323
307
536
577
.6619
661
703
530
520
1.511
501
492
195
197
1.199
202
204
828
820
1.812
804
796
56 50'
40'
30'
20'
10'
34 00'
.5592
.8290
.6745
1.483
1.206
1.788
56 00'
10'
20'
30
40
34 50
616
640
.5664
688
712
274
258
.8241
225
208
787
830
.6873
916
.6959
473
464
1.455
446
437
209
211
1.213
216
218
781
773
1.766
758
751
55 50'
40'
30'
20'
10'
35 00'
.5736
.8192
.7002
1.428
1.221
1.743
55 00'
10'
20'
30'
40'
35 50'
760
783
.5807
831
854
175
158
.8141
124
107
046
089
.7133
177
221
419
411
1.402
393
385
223
226
1.228
231
233
736
729
1.722
715
708
54 50'
40'
30'
20'
10'
36 00'
.5878
.8090
.7265
1.376
1.236
1.701
54 00'
Cos
Sin
Cot
Tan
Csc
Sec
-<
Appencf/x A
319
TABLE II (continued)
-
Sin
Cos
Tan
Cot
Sec
Csc
36 W
.5878
.8090
.7265
1.376
1.236
1.701
54 Q 00'
10'
20'
30'
40'
36 50*
901
925
.5948
972
.5995
073
056
.8039
021
.8004
310
355
.7400
445
490
368
360
1.351
343
335
239
241
1.244
247
249
695
688
1.681
675
668
53 50'
40'
30'
20'
10'
3T00'
.6018
.7986
.7536
1.327
1.252
1.662
53 00'
10'
20'
30'
40'
37 50'
041
065
.6088
111
134
969
951
.7934
916
898
581
627
.7673
720
766
319
311
1.303
295
288
255
258
1.260
263
266
655
649
1.643
636
630
52 50'
40'
30'
20'
10'
38 00'
.6157
.7880
.7813
1.280
1.269
1.624
62 00'
10'
20'
30'
40'
38 50'
180
202
.6225
248
271
862
844
.7826
808
790
860
907
.7954
.8002
050
272
265
1.257
250
242
272
275
1.278
281
284
618
612
1.606
601
595
51 50'
40'
30'
20'
10'
39 00'
.6293
.7771
.8098
1.235
1.287
1.589
51 00'
10'
20'
30'
40'
39 50'
316
338
.6361
383
406
753
735
.7716
698
679
146
195
.8243
292
342
228
220
1.213
206
199
290
293
1.296
299
302
583
578
1.572
567
561
50 50'
40'
30'
20'
10'
40 00'
.6428
.7660
.8391
1.192
1.305
1.556
50 00
10
20'
30'
40'
40 50'
450
472
.6494
517
539
642
623
.7604
585
566
441
491
.8541
591
642
185
178
1.171
164
157
309
312
1.315
318
322
550
545
1.540
535
529
49 50'
40'
30'
20'
10'
41 00'
.6561
.7547
.8693
1.150
1.325
1.524
49 00'
10'
20'
30'
40'
41 50'
583
604
.6626
648
670
528
509
.7490
470
451
744
796
.8847
899
.8952
144
137
1.130
124
117
328
332
1.335
339
342
519
514
1.509
504
499
48 50'
40'
30'
20'
10'
42 00'
.6691
.7431
.9004
1.111
1.346
1.494
48 00'
Cos
Sin
Cot
Tan
Csc
Sec
4
320
Appendix A
TABLE II (continued)
I"""'"
Sin
Cos
Tan
Cot
Sec
Csc
42 00'
.6691
.7431
.9004
1.111
1.346
1.494
48 00'
10'
20'
30'
40'
42 60'
713
734
.6756
777
799
412
392
.7373
353
333
057
110
.9163
217
271
104
098
1.091
085
079
349
353
1.356
360
364
490
485
1.480
476
471
47 50'
40'
30'
20'
1(X
43 00'
.6820
.7314
.9325
1.072
1.367
1.466
47 00'
10'
20'
30'
40'
43 50'
841
862
.6884
905
926
294
274
.7254
234
214
380
435
.9490
545
601
066
060
1.054
048
042
371
375
1.379
382
386
462
457
1.453
448
444
46 50'
40'
30'
20'
10'
44 00*
.6947
.7193
.9657
1.036
1.390
1.440
46 00'
10'
20'
30*
40'
44 60'
967
.6988
.7009
030
050
173
153
.7133
112
092
713
770
.9827
884
.9942
030
024
1.018
012
006
394
398
1.402
406
410
435
431
1.427
423
418
45 50'
40'
30'
20'
10'
uw
.7071
.7071
1.000
1.000
1.414
1.414
45 00'
Cos
Sin
Cot
Tan
Cse
See
<
Appendix A
32V
TABLE III
FOUR-PLACE LOGARITHMS OF NUMBERS
W
u
10
.0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
11
12
13
.0414
.0792
.1139
0453
0828
1173
0492
0864
1206
0531
0899
1239
0569
0934
1271
0607
0969
1303
0645
1004
1335
0682
1038
1367
0719
1072
1399
0755
1106
1430
14
15
16
.1461
.1761
.2041
1492
1790
2068
1523
1818
2095
1553
1847
2122
1584
1875
2148
1614
1903
2175
1644
1931
2201
1673
1959
2227
1703
1987
2253
1732
2014
2279
17
18
19
.2304
.2553
.2788
2330
2577
2810
2355
2601
2833
2380
2625
2856
2405
2648
2878
2430
2672
2900
2455
2695
2923
2480
2718
2945
2504
2742
2967
2529
2765
2989
20
.3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
22
23
.3222
.3424
.3617
3243
3444
3636
3263
3464
3655
3284
3483
3674
3304
3502
3692
3324
3522
3711
3345
3541
3729
3365
3560
3747
3385
3579
3766
3404
3598
3784
24
25
26
.3802
.3979
.4150
3820
3997
4166
3838
4014
4183
3856
4031
4200
3874
4048
4216
3892
4065
4232
3909
4082
4249
3927
4099
4265
3945
4116
4281
3962
4133
4298
27
28
29
.4314
.4472
.4624
4330
4487
4639
4346
4502
4654
4362
4518
4669
4378
4533
4683
4393
4548
4698
4409
4564
4713
4425
4579
4728
4440
4594
4742
4456
4609
4757
30
.4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
32
33
.4914
.5051
.5185
4928
5065
5198
4942
5079
5211
4955
5092
5224
4969
5105
5237
4983
5119
5250
4997
5132
5263
5011
5145
5276
5024
5159
5289
5038
5172
5302
34
35
36
.5315
.5441
.5563
5328
5453
5575
5340
5465
5587
5353
5478
5599
5366
5490
5611
5378
5502
5623
5391
5514
5635
5403
5527
5647
5416
5539
5658
5428
5551
5670
37
38
39
.5682
.5798
.5911
5694
5809
5922
5705
5821
5933
5717
5832
5944
5729
5843
5955
5740
5855
5966
5752
5866
5977
5763
5877
5988
5775
5888
5999
5786
5899
6010
40
.6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
N
1
2
3
4
5
6
7
8
9
322
Appendix A
TABLE III (continued)
N
1
3
3
4
5
6
7
8
9
40
.6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
41
42
43
.6128
.6232
.6335
6138
6243
6345
6149
6253
6355
6160
6263
6365
6170
6274
6375
6180
5284
6385
6191
6294
6395
6201
6304
6405
6212
6314
6415
6222
6325
6425
44
48
46
.6435
.6532
.6628
6444
6542
6637
6454
6551
6646
6464
6561
6656
6474
6571
6665
6484
6580
6675
6493
6590
6684
6503
6599
6693
6513
6609
6702
6522
6618
6712
47
48
49
.6721
.6812
.6902
6730
6821
6911
6739
6830
6920
6749
6839
6928
6758
6848
6937
6767
6857
6946
6776
6866
6955
6785
6875
6964
6794
6884
6972
6803
6893
6981
60
.6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
52
53
.7076
.7160
.7243
7084
7168
7251
7093
7177
7259
7101
7185
7267
7110
7193
7275
7118
7202
7284
7126
7210
7292
7135
7218
7300
7143
7226
7308
7152
7235
7316
54
55
56
.7324
.7404
.7482
7332
7412
7490
7340
7419
7497
7348
7427
7505
7356
7435
7513
7364
7443
7520
7372
7451
7528
7380
7459
7536
7388
7466
7543
7396
7474
7551
57
58
59
.7559
.7634
.7709
7566
7642
7716
7574
7649
7723
7582
7657
7731
7589
7664
7738
7597
7672
7745
7604
7679
7752
7612
7686
7760
7619
7694
7767
7627
7701
7774
60
.7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
62
63
.7853
.7924
.7993
7860
7931
8000
7868
7938
8007
7875
7945
8014
7882
7952
8021
7889
7959
8028
7896
7966
8035
7903
7973
8041
7910
7980
8048
7917
7987
8055
64
65
66
.8062
.8129
.8195
8069
8136
8202
8075
8142
8209
8082
8149
8215
8089
8156
8222
8096
8162
8228
8102
8169
8235
8109
8176
8241
8116
8182
8248
8122
8189
8254
67
68
69
.8261
.8325
.8388
8267
8331
8395
8274
8338
8401
8280
8344
8407
8287
8351
8414
8293
8357
8420
8299
8363
8426
8306
8370
8432
8312
8376
8439
8319
8382
8445
70
.8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
H
1
*
3
4
5
6
7
8
Appendix A
323
TABLE III (continued)
N
70
71
72
73
1
S
3
4
5
6
7
8
9
.8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
.8513
.8573
.8633
8519
8579
8639
8525
8585
8645
8531
8591
8651
8537
8597
8657
8543
8603
8663
8549
8609
8669
8555
8615
8675
8561
8621
8681
8567
8627
8686
74
75
76
.8692
.8751
.8808
8698
8756
8814
8704
8762
8820
8710
8768
8825
8716
8774
8831
8722
8779
8837
8727
8785
8842
8733
8791
8848
8739
8797
8854
8745
8802
8859
77
78
79
80
.8865
.8921
.8976
8871
8927
8982
8876
8932
8987
8882
8938
8993
8887
8943
8998
8893
8949
9004
8899
8954
9009
8904
8960
9015
8910
8965
9020
8915
8971
9025
.9031
9036
9042
9047
9053
9058
9063
9069
9074
9128
9180
9232
9079
9133
9186
9238
81
82
83
.9085
.9138
.9191
9090
9143
9196
9096
9149
9201
9101
9154
9206
9106
9159
9212
9112
9165
9217
9117
9170
9222
9122
9175
9227
84
85
86
.9243
.9294
.9345
9248
9299
9350
9253
9304
9355
9258
9309
9360
9263
9315
9365
9269
9320
9370
9274
9325
9375
9279
9330
9380
9284
9335
9385
9289
9340
9390
87
88
89
.9395
.9445
.9494
9400
9450
9499
9405
9455
9504
9410
9460
9509
9415
9465
9513
9420
9469
9518
9425
9474
9523
9430
9479
9528
9435
9484
9533
9440
9489
9538
90
.9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
91
92
93
.9590
.9638
.9685
9595
9643
9689
9600
9647
9694
9605
9652
9699
9609
9657
9703
9614
9661
9708
9619
9666
9713
9624
9671
9717
9628
9675
9722
9633
9680
9727
94
95
96
.9731
.9777
.9823
9736
9782
9827
9741
9786
9832
9745
9791
9836
9750
9795
9841
9754
9800
9845
9759
9805
9850
9763
9809
9854
9768
9814
9859
9773
9818
9863
97
98
99
.9868
.9912
.9956
9872
9917
9961
9877
9921
9965
9881
9926
9969
9886
9930
9974
9890
9934
9978
9894
9939
9983
9899
9943
9987
9903
9948
9991
9908
9952
9996
N
1
2
3
4
5
6
7
8
9
324
Appendix A
TABLE IV
FOUR-PLACE LOGARITHMS OF FUNCTIONS
>-
LSin
L Tan
LCot
L Cos
000'
10.0000
90 00'
10'
20'
30'
40'
050'
7.4637
.7648
7.9408
8.0658
.1627
7.4637
.7648
7.9409
8.0658
.1627
12.5363
.2352
12.0591
11.9342
.8373
.0000
.0000
.0000
.0000
10.0000
89 50'
40'
30'
20'
10'
1<> 00'
8.2419
8.2419
11.7581
9.9999
89* 00'
10'
20'
30'
40'
1 50'
.3088
.3668
.4179
.4637
.5050
.3089
.3669
.4181
.4638
.5053
.6911
.6331
.5819
.5362
.4947
.9999
.9999
.9999
.9998
.9998
88 50'
40'
30'
20'
10'
2 00'
8.5428
8.5431
11.4569
9.9997
88 00'
10'
20'
30'
40'
2 50'
.5776
.6097
.6397
.6677
.6940
.5779
.6101
.6401
.6682
.6945
.4221
.3899
.3599
.3318
.3055
.9997
.9996
.9996
.9995
.9995
87 50'
40'
30'
20'
10'
3 00'
8.7188
8.7194
11.2806
9.9994
87 00'
10'
20'
30'
40'
3 50'
.7423
.7645
.7857
.8059
.8251
.7429
.7652
.7865
.8067
.8261
.2571
.2348
.2135
.1933
.1739
.9993
.9993
.9992
.9991
.9990
86 50'
40'
30'
20'
10'
4 00'
8.8436
8.8446
11.1554
9.9989
86 00'
10'
20'
30'
40'
4 50'
.8613
.8783
.8946
.9104
.9256
.8624
.8795
.8960
.9118
.9272
.1376
.1205
.1040
.0882
.0728
.9989
.9988
.9987
.9986
.9985
85 50'
40'
30'
20'
10'
5 00'
8.9403
8.9420
11.0580
9.9983
85 00'
10'
20'
30'
40'
5 50'
.9545
.9682
.9816
8.9945
9.0070
.9563
.9701
.9836
8.9966
9.0093
.0437
.0299
.0164
11.0034
10.9907
.9982
.9981
.9980
.9979
.9977
84 50'
40'
30'
20'
10'
6 00'
9.0192
9.0216
10.9784
9.9976
84 00'
L Cos
L Cot
L Tan
LSin
*
Appendix A
325
TABLE IV (continued)
*
L Sin
LTan
L Cot
L Cos
6 00'
9.0192
9.0216
10.9784
9.9976
84 00'
10'
20'
30'
40'
6 50'
.0311
.0426
.0539
.0648
.0755
.0336
.0453
.0567
.0678
.0786
.9664
.9547
.9433
.9322
.9214
.9975
.9973
.9972
.9971
.9969
83 50'
40'
30'
20'
10'
7 00'
9.0859
9.0891
10.9109
9.9968
83 00'
10'
20'
30'
40'
7 50'
.0961
.1060
.1157
.1252
.1345
.0995
.1096
.1194
.1291
.1385
.9005
.8904
.8806
.8709
.8615
.9966
.9964
.9963
.9961
.9959
82 50'
40'
30'
20'
10'
8 00'
9.1436
9.1478
10.8522
9.9958
82 00'
10'
20'
30'
40'
8 50'
.1525
.1612
.1697
.1781
.1863
.1569
.1658
.1745
.1831
.1915
.8431
.8342
.8255
.8169
.8085
.9956
.9954
.9952
.9950
.9948
81 50'
40'
30'
20'
10'
9 00'
9.1943
9.1997
10.8003
9.9946
81 00'
10'
20'
30'
40'
9 50'
.2022
.2100
.2176
.2251
.2324
.2078
.2158
.2236
.2313
.23.89
.7922
.7842
.7764
.7687
.7611
.9944
.9942
.9940
.9938
.9936
80 50'
40'
30'
20'
10'
10 00'
9.2397
9.2463
10.7537
9.9934
80 00'
10'
20'
30'
40'
10 50'
.2468
.2538
.2606
.2674
.2740
.2536
.2609
.2680
.2750
.2819
.7464
.7391
.7320
.7250
.7181
.9931
.9929
.9927
.9924
.9922
79 50'
40'
30'
20'
10'
11 00'
9.2806
9.2887
10.7113
9.9919
79 00'
10'
20'
30'
40'
11 50'
.2870
.2934
.2997
.3058
.3119
.2953
.3020
.3085
.3149
.3212
.7047
.6980
.6915
.6851
.6788
.9917
.9914
.9912
.9909
.9907
78 50'
40'
30'
20'
10'.
1S 00'
9.3179
9.3275
10.6725
9.9904
78 00'
L Cos
L Cot
LTan
L Sin
326
Appendix A
TABLE IV (continued)
*
LSin
LTan
LCot
LCos
12 00'
9.3179
9.3275
10.6725
9.9904
78 00'
10'
20'
30'
40'
12 50'
.3238
.3296
.3353
.3410
.3466
.3336
.3397
.3458
.3517
.3576
.6664
.6603
.6542
.6483
.6424
.9901
.9899
.9896
.9893
.9890
77 50'
40'
30'
20'
10'
13 00'
9.3521
9.3634
10.6366
9.9887
77 00'
10'
20'
30'
40'
13 50'
.3575
.3629
.3682
.3734
.3786
.3691
.3748
.3804
.3859
.3914
.6309
.6252
.6196
.6141
.6086
.9884
.9881
.9878
.9875
.9872
76 50'
40'
30'
20'
10'
14 00'
9.3837
9.3968
10.6032
9.9869
76 00'
10'
20'
30'
40'
14 50'
.3887
.3937
.3986
.4035
.4083
.4021
.4074
.4127
.4178
.4230
.5979
.5926
.5873
.5822
.5770
.9866
.9863
.9859
.9856
.9853
75 50'
40'
30'
20'
10'
15 00'
9.4130
9.4281
10.5719
9.9849
75 00'
10'
20'
30'
40'
15 50'
.4177
.4223
.4269
.4314
.4359
.4331
.4381
.4430
.4479
.4527
.5669
.5619
.5570
.5521
.5473
.9846
.9843
.9839
.9836
.9832
74 50'
40'
30'
20'
10'
16 00'
9.4403
9.4575
10.5425
9.9828
74 00'
10'
20'
30'
40'
16 50'
.4447
.4491
.4533
.4576
.4618
.4622
.4669
.4716
.4762
.4808
.5378
.5331
.5284
.5238
.5192
.9825
.9821
.9817
.9814
.9810
73 50'
40'
30'
20'
10'
17 00'
9.4659
9.4853
10.5147
9.9806
73 00'
10'
20'
30'
40'
17 50'
.4700
.4741
.4781
.4821
.4861
.4898
.4943
.4987
.5031
.5075
.5102
.5057
.5013
.4969
.4925
.9802
.9798
.9794
.9790
.9786
72 50'
40'
30'
20'
10'
18 00'
9.4900
9.5118
10.4882
9.9782
72 00'
LCos
LCot
LTan
LSin
<
Appendix A
327
TABLE IV (continued)
*
LSin
LTan
LCot
LCos
18 00'
9.4900
9.5118
10.4882
9.9782
72 00'
10'
20'
30'
40'
18 50'
.4939
.4977
.5015
.5052
.5090
.5161
.5203
.5245
.5287
.5329
.4839
.4797
.4755
.4713
.4671
.9778
.9774
.9770
.9765
.9761
71 50'
40'
30'
20'
10'
19 00'
9.5126
9.5370
10.4630
9.9757
71 00'
10'
20'
30'
40'
19 50'
.5163
.5199
.5235
.5270
.5306
.5411
.5451
.5491
.5531
.5571
.4589
.4549
.4509
.4469
.4429
.9752
.9748
.9743
.9739
.9734
70 50'
40'
30*
20'
10'
20 00'
9.5341
9.5611
10.4389
9.9730
70 W
10'
20'
30'
40'
20 50'
.5375
.5409
.5443
.5477
.5510
.5650
.5689
.5727
.5766
.5804
.4350
.4311
.4273
.4234
.4196
.9725
.9721
.9716
.9711
.9706
69 50'
40'
30'
20'
10*
21 00'
9.5543
9.5842
10.4158
9.9702
69 00*
10'
20'
30'
40'
21 50'
.5576
.5609
.5641
.5673
.5704
.5879
.5917
.5954
.5991
.6028
.4121
.4083
.4046
.4009
.3972
.9697
.9692
.9687
.9682
.9677
68 W
4<y
30*
20'
10'
22 00'
9.5736
9.6064
10.3936
9.9672
68 00'
10'
20'
30'
40'
22 50'
.5767
.5798
.5828
.5859
.5889
.6100
.6136
.6'172
.0208
.6243
.3900
.3864
.3828
.3792
.3757
.9667
.9661
.9656
.9651
.9646
67 50'
40'
30'
20'
10'
23 00'
9.5919
9.6279
10.3721
9.9640
67 00'
10'
20'
30'
40'
23 50'
.5948
.5978
.6007
.6036
.6065
.6314
.6348
.6383
.6417
.6452
.3686
.3652
.3617
.3583
.3548
.9635
.9629
.9624
.9618
.9613
66 50'
40'
30'
20'
10'
24 00'
9.6093
9.6486
10.3514
9.9607
66 00'
L Cos
LCot
LTan
LSin
*
328
Appendix A
TABLE IV (continued)
.
L Sin (
LTan
LCot
L Cos
24 00'
9.6093
9.6486
10.3514
9.9607
66 00'
10'
20'
30'
40'
24 50'
.6121
.6149
.6177
.6205
.6232
.6520
.6553
.6587
.6620
.6654
.3480
.3447
.3413
.3380
.3346
.9602
.9596
.9590
.9584
.9579
65 50'
40'
30'
20'
10'
25 00'
9.6259
9.6687
10.3313
9.9573
65 00'
10'
20'
so'
40'
25 50'
.6286
.6313
.6340
.6366
.6392
.6720
.6752
.6785
.6817
.6850
.3280
.3248
.3215
.3183
.3150
.9567
.9561
.9555
.9549
.9543
64 50'
40'
30'
20'
10'
26 00'
9.6418
9.6882
10.3118
9.9537
64 00'
10'
20'
30'
40'
26 50'
.6444
.6470
.6495
.6521
.6546
.6914
.6946
.6977
.7009
.7040
.3086
.3054
.3023
.2991
.2960
.9530
.9524
.9518
.9512
.9505
63 50'
40'
30'
20'
10'
27 00'
9.6570
9.7072
10.2928
9.9499
63 00'
10'
20'
30'
40'
27 50'
.6595
.6620
.6644
.6668
.6692
.7103
.7134
.7165
.7196
.7226
.2897
.2866
.2835
.2804
.2774
.9492
.9486
.9479
.9473
.9466
62 50'
40'
30'
20'
10'
28 00'
9.6716
9.7257
10.2743
9.9459
62 00'
10'
20'
30'
40'
28 50'
.6740
.6763
.6787
.6810
.6833
.7287
.7317
.7348
.7378
.7408
.2713
.2683
.2652
.2622
.2592
.9453
.9446
.9439
.9432
.9425
61 50'
40'
30'
20'
10'
29 00'
9.6856
9.7438
10.2562
9.9418
61 00'
10'
20'
30'
40'
29 50'
.6878
.6901
.6923
.6946
.6968
.7467
.7497
.7526
.7556
.7585
.2533
.2503
.2474
.2444
.2415
.9411
.9404
.9397
.9390
.9383
60 50'
40'
30'
W
W
30 00'
9.6990
9.7614
10.2386
9.9375
60 00'
LCos
LCot
LTan
LSin
4
Appendix A
TABLE IV (continued)
329
*
L Sin
LTan
LCot
LCos
30 00'
9.6990
9.7614
10.2386
9.9375
W00'
10'
20'
30'
40'
30 50'
.7012
.7033
.7055
.7076
.7097
.7644
.7673
.7701
.7730
.7759
.2356
.2327
.2299
.2270
.2241
.9368
.9361
.9353
.9346
.9338
59 50'
40'
30'
20'
10'
31 00'
9.7118
9.7788
10.2212
9.9331
59 00'
10'
20'
30'
40'
31 50'
.7139
.7160
.7181
.7201
.7222
.7816
.7845
.7873
.7902
.7930
.2184
.2155
.2127
.2098
.2070
.9323
.9315
.9308
.9300
.9292
58 50'
40'
30'
20'
10'
32 00'
9.7242
9.7958
10.2042
9.9284
58 00'
10'
20'
30'
40'
32 50'
.7262
.7282
.7302
.7322
.7342
.7986
.8014
.8042
.8070
.8097
.2014
.1986
.1958
.1930
.1903
.9276
.9268
.9260
.9252
.9244
57 50'
40'
30'
20'
10'
33 00'
9.7361
9.8125
10.1875
9.9236
57 00'
10'
20'
30'
40'
33 50'
.7380
.7400
.7419
.7438
.7457
.8153
.8180
.8208
.8235
.8263
.1847
.1820
.1792
.1765
.1737
.9228
.9219
.9211
.9203
.9194
56 50'
40'
30'
20'
10'
34 00'
9.7476
9.8290
10.1710
9.9186
56 00'
10'
20'
30'
40'
34 50'
.7494
.7513
.7531
.7550
.7568
.8317
.8344
.8371
.8398
.8425
.1683
.1656
.1629
.1602
.1575
.9177
.9169
.9160
.9151
.9142
55 50'
40'
30'
20'
10'
35 00'
9.7586
9.8452
10.1548
9.9134
55 00'
10'
20'
30'
40'
35 50'
.7604
.7622
.7640
.7657
.7675
.8479
.8506
.8533
.8559
.8586
.1521
.1494
.1467
.1441
.1414
.9125
.9116
.9107
.9098
.9089
54 50'
40'
30'
20'
10'
36 00'
9.7692
9.8613
10.1387
9.9080
64 00'
LCos
LCot
LTan
LSin
-
330
Appendix A
TABLE IV (continued)
*.
L Sin
LTan
LCot
LCos
36 00'
9.7692
9.8613
10.1387
9.9080
64 00'
10'
20'
30'
40'
36 50'
.7710
.7727
.7744
.7761
.7778
.8639
.8666
.8692
.8718
.8745
.1361
.1334
.1308
.1282
.1255
.9070
.9061
.9052
.9042
.9033
53 50'
40'
30'
20'
10'
37 00'
9.7795
9.8771
10.1229
9.9023
53 00'
10'
20'
30'
40'
37 50'
.7811
.7828
.7844
.7861
.7877
.8797
.8824
.8850
.8876
.8902
.1203
.1176
.1150
.1124
.1098
.9014
.9004
.8995
.8985
.8975
52 50'
40'
30'
20'
10'
38 00'
9.7893
9.8928
10.1072
9.8965
52 00'
10'
20'
30'
40'
38 50'
.7910
.7926
.7941
.7957
.7973
.8954
.8980
.9006
.9032
.9058
.1046
.1020
.0994
.0968
.0942
.8955
.8945
.8935
.8925
.8915
51 50'
40'
30'
20'
10'
39 00'
9.7989
9.9084
10.0916
9.8905
51 00'
10'
20'
30'
40'
39 50'
.8004
.8020
.8035
.8050
.8066
.9110
.9135
.9161
.9187
.9212
.0890
.0865
.0839
.0813
.0788
.8895
.8884
.8874
.8864
.8853
50 50'
40'
30'
20'
10'
40 00'
9.8081
9.9238
10.0762
9.8843
50 00'
10'
20'
30'
40'
40 50'
.8096
.8111
.8125
.8140
.8155
.9264
.9289
.9315
.9341
.9366
.0736
.0711
.0685
.0659
.0634
.8832
.8821
.8810
.8800
.8789
49 50'
40'
30'
20'
10'
41 00'
9.8169
9.9392
10.0608
9.8778
49 00'
10'
20'
30'
40'
41 50'
.8184
.8198
.8213
.8227
.8241
.9417
.9443
.9468
.9494
.9519
.0583
.0557
.0532
.0506
.0481
.8767
.8756
.8745
.8733
.8722
48 50'
40'
30'
20'
10'
42 00'
9.8255
9.9544
10.0456
9.8711
48 00'
LCos
LCot
LTan
LSin
^
Appendix A
331
TABLE IV (continued)
>-
L Sin
L Tan
LCot
L Cos
42 00'
9.8255
9.9544
10.0456
9.8711
48 00'
10'
20'
30'
40'
42 50'
.8269
.8283
.8297
.8311
.8324
.9570
.9595
.9621
.9646
.9671
.0430
.0405
.0379
.0354
.0329
.8699
.8688
.8676
.8665
.8653
47 50'
40'
30'
20'
10'
43 00'
9.8338
9.9697
10.0303
9.8641
47 00'
10'
20'
30'
40'
43 50'
.8351
.8365
.8378
.8391
.8405
.9722
.9747
.9772
.9798
.9823
.0278
.0253
.0228
.0202
.0177
.8629
.8618
.8606
.8594
.8582
46 50'
40'
30'
20'
10'
44 00'
9.8418
9.9848
10.0152
9.8569
46 00'
10'
20'
30'
40'
44 50'
.8431
.8444
.8457
.8469
.8482
.9874
.9899
.9924
.9949
.9975
.0126
.0101
.0076
.0051
.0025
.8557
.8545
.8532
.8520
.8507
45 50'
40'
30'
20'
10'
45 00'
9.8495
10.0000
10.0000
9.8495
45 00'
L Cos
LCot
LTan
LSin
*
332
Appendix A
TABLE V
SQUARES AND SQUARE ROOTS
H
JV*
VN
VlON
JV
N*
VN
VlON
N
N*
VN
ViQN
1.00
1.0000
1.00000
3.16228
1.60
2.5600
1.26491
4.00000
2.20
4.8400
1.48324
4.69042
1.01
1.0201
1.00499
3.17805
1.61
2.5921
1.26886
4.01248
2.21
4.8841
1.48661
4.70106
1.02
1.0404
1.00995
3.19374
1.62
2.6244
1.27279
4.02492
2.22
4.9284
1.48997
4.71169
1.03
1.0609
1.01489
3.20936
1.63
2.6669
1.27671
4.03733
2.23
4.9729
1.49332
4.72229
1.04
1.0816
1.01980
3.22490
1.64
2.6896
1.28062
4.04969
2.24
5.0176
1.49666
4.73286
1.05
1.1025
1.02470
3.24037
1.65
2.7225
1.28462
4.06202
2.25
5.0625
1.60000
4.74342
1.06
1.1236
1.02956
3.25576
1.66
2.7556
1.28841
4.07431
2.26
5.1076
1.60333
4.75395
1.07
1.1449
1.03441
3.27109
1.67
2.7889
1.29228
4.08656
2.27
5.1529
1.50665
4.76445
1.08
1.1664
1.03923
3.28634
1.68
2.8224
1.29615
4.09878
2.28
5.1984
1.50997
4.77493
1.09
1.1881
1.04403
3.30151
1.69
2.8661
1.30000
4.11096
2.29
5.2441
1.51327
4.78539
1.10
1.2100
1.04881
3.31662
1.70
2.8900
1.30384
4.12311
2.30
'5.2900
1.51668
4.79583
1.11
1.2321
1.05357
3.33167
1.71
2.9241
1.30767
4.13521
2.31
5.3361
1.51987
4.80625
1.12
1.2544
1.05830
3.34664
1.72
2.9684
1.31149
4.14729
2.32
6.3824
1.62315
4.81664
1.13
1.2769
1.06301
3.36155
1.73
2.9929
1.31629
4.15933
2.33
5.4289
1.52643
4.82701
1.14
1.2996
1.06771
3.37639
1.74
3.0276
1.31909
4.17133
2.34
6.4756
1.52971
4.83735
1.15
1.3225
1.07238
3.39116
1.75
3.0625
1.32288
4.18330
2.35
6.5225
1.63297
4.84768
1.16
1.3456
1.07703
3.40588
1.76
3.0976
1.32666
4.19524
2.36
5.5696
1.63623
4.85798
1.17
1.3689
1.08167
3.42053
1.77
3.1329
1.33041
4.20714
2.37
66169
1.53948
4.86826
1.18
1.3924
1.08628
3.43511
1.78
3.1684
1.33417
4.21900
2.38
5.6644
1.54272
4.87852
1.19
1.4161
1.09087
3.44964
1.79
3.2041
1.33791
4.23084
2.39
6.7121
1.64596
4.88876
l.*0
1.4400
1.09545
3.46410
1.80
3.2400
1.34164
4.24264
2.40
5.7600
1.64919
4.89898
1.21
1.4641
.10000
3.47851
1.81
3.2761
1.34636
4.26441
2.41
6.8081
1.56242
4.90918
1.22
1.4884
.10454
3.49285
1.82
3.3124
1.34907
4.26615
2.42
5.8564
1.65563
4.91935
1.23
1.5129
.10905
3.50714
1.83
3.3489
1.36277
4.27785
2.43
5.9049
1.55885
4.92950
1.24
1.5376
.11355
3.52136
1.84
3.3856
1.35647
4.28952
2.44
6.9536
1.56205
4.93964
1.15
1.5625
.11803
3.53553
1.85
3.4225
1.36015
4.30116
2.45
6.0025
1.56525
4.94975
1.26
1.6876
.12250
3.54965
1.86
3.4696
1.36382
4.31277
2.46
6.0516
1.66844
4.96984
1.27
1.6129
.12694
3.56371
1.87
3.4969
1.36748
4.32435
2.47
6.1009
1.57162
4.96991
1.28
1.6384
.13137
3.57771
1.88
3.6344
1.37113
4.33590
2.48
6.1604
1.57480
4.97996
1.29
1.6641
.13578
3.59166
1.89
3.5721
1.37477
4.34741
2.49
6.2001
1.57797
4.98999
1.80
1.6900
.14018
3.60555
1.90
3.6100
1.37840
4.36890
2.50
6.2500
1.58114
6.00000
1.31
1.7161
.14455
3.61939
1.91
3.6481
1.38203
4.37035
2.51
6.3001
1.58430
5.00999
1.32
1.7424
.14891
3.63318
1.92
3.6864
1.38664
4.38178
2.52
6.3504
1.58745
5.01996
1.33
1.7689
.15326
3.64692
1.93
37249
1.38924
4.39318
2.53
6.4009
1.59060
5.02991
1.34
1.7956
.15758
3.66060
1.94
3.7036
1.39284
4.40454
2.54
6.4516
1.69374
6.03984
1.35
1.8225
1.16190
3.67423
1.95
3.8026
1.39642
4.41588
2.55
6.5025
1.59687
5.04975
1.36
1.8496
1.16619
3.68782
1.96
3.8416
1.40000
4.42719
2.56
6.5536
1.60000
5.05964
1.37
1.8769
1.17047
3.70135
1.97
3.8809
1.40357
4.43847
2.57
6.6049
1.60312
5.06952
1.38
1.9044
1.17473
3.71484
1.98
3.9204
1.40712
4.44972
2.58
6.6564
1.60624
5.07937
1.39
1.9321
1.17898
3.72827
1.99
3.9601
1.41067
4.46094
2.59
6.7081
1.60935
5.08920
1.40
1.9600
1.18322
3.74166
2.00
4.0000
1.41421
4.47214
2.60
6.7600
1.61245
6.09902
1.41
1.9881
1.18743
3.75500
2.01
4.0401
1.41774
4.48330
2.61
6.8121
1.61566
5.10882
1.42
2.0164
1.19164
3.76829
2.02
4.0804
1.42127
4.49444
2.62
6.8644
.61864
6.11859
1.43
2.0449
1.19583
3.78163
2.03
4.1209
1.42478
4.60555
2.63
6.9169
.62173
6.12835
1.44
2.0736
1.20000
3.79473
2.04
4.1616
1.42829
4.51664
2.64
6.9696
.62481
5.13809
1.45
2.1025
1.20416
3.80789
2.05
4.2025
1.43178
4.52769
2.65
7.0225
1.62788
6.14782
1.46
2.1316
1.20830
3.82099
2.06
4.2436
1.43527
4.53872
2.66
7.0756
.63095
5.15752
1.47
2.1609
1.21244
3.83406
2.07
4.2849
1.43875
4.64973
2.67
7.1289
.63401
5.16720
1.48
2.1904
1.21655
3.84708
2.08
4.3264
1.44222
4.56070
2.68
7.1824
.63707
6.17687
1.49
2.2201
1.22066
3.86005
2.09
4.3681
1.44668
4.57165
2.69
7.2361
1.64012
5.18652
1.60
2.2500
1.22474
3.87293
S.10
4.4100
1.44914
4.58258
2.70
7.2900
1.64317
5.19615
1.51
2.2801
1.22882
3.88587
2.11
4.4521
1.45258
4.59347
2.71
7.3441
1.64621
5.20577
1.52
2.3104
1.23288
3.89872
2.12
4.4944
1.45602
4.60435
2.72
7.3984
1.64924
5.21636
1.53
2.3409
1.23693
3.91152
2.13
4.5369
1.45945
4.61519
2.73
7.4529
1.65227
5.22494
1.54
2.3716
1.24097
3.92428
2.14
4.5796
1.46287
4.62601
2.74
7.5076
1.65529
5.23450
1.55
2.4025
1.24499
3.93700
S.15
4.6225
1.46629
4.63681
2.75
7.5626
1.66831
5.24404
1.56
2.4336
1.24900
3.94968
2.16
4.6656
1.46969
4.64758
2.76
7.6176
1.66132
5.26357
1.57
2.4649
1.25300
3.96232
2.17
4.7089
1.47309
4.66833
2.77
7.6729
1.66433
6.26308
1.58
2.4964
1.25698
3.97492
2.18
4.7624
1.47648
4.66905
2.78
7.7284
1.66733
5.27257
1.59
2.5281
1.26095
3.98748
2.19
4.7961
1.47986
4.67974
2.79
7.7841
1.67033
5.28205
MO
2.5600
1.26491
4.00000
MO
4.8400
1.48324
4.69042
2.80
7.8400
1.67332
5.29160
N
JV
VN
vlo??
N
N*
VN
VIM
N
- AP
VN
View
Appendix A
333
TABLE V (continued)
H
N*
VN
viov
N
N 9
VN
viSf
N
N*
VN
vI5y
1.80
7.8400
1.67332
5.29150
8.40
11.5600
1.84391
6.83095
4.00
16.0000
2.00000
6.32456
2.81
7.8961
1.67631
5.30094
3.41
11.6281
1.84662
5.83952
4.01
16.0801
2.00250
6.33246
2.82
7.9524
1.67929
5.31037
3.42
11.6964
1.84932
5.84808
4.02
16.1604
2.00499
6.34035
2.83
8.0089
1.68226
5.31977
3.43
11.7649
1.85203
5.85662
4.03
16.2409
2.00749
6.34823
2.84
8.0656
1.68523
5.32917
3.44
11.8336
1.85472
5.86515
4.04
16.3216
2.00998
6.35610
1.89
8.1225
1.68819
5.33854
8.49
11.9025
1.85742
5.87367
4.05
16.4025
2.01246
6.36396
2.86
8.1796
1.69115
5.34790
3.46
11.9716
1.86011
5.88218
4.06
16.4836
2.01494
6.37181
2.87
8.2369
1.69411
5.35724
3.47
12.0409
1.86279
5.89067
4.07
16.5649
2.01742
6.37966
2.88
8.2944
1.69706
5.36656
3.48
12.1104
1.86548
5.89915
4.08
16.6464
2.01990
6.38749
2.89
8.3521
.70000
5.37587
3.49
12.1801
1.86815
5.90762
4.09
16.7281
2.02237
6.39531
1.90
8.4100
70294
5.38516
3.80
12.2500
1.87083
5.91608
4.10
16.8100
2.02485
6.40312
2.91
8.4681
.70587
5.39444
3.51
12.3201
1.87350
5.92453
4.11
16.8921
2.02731
6.41093
2.92
8.5264
.70880
5.40370
3.52
12.3904
1.87617
5.9&S6
4.12
16.9744
2.02978
6.41872
2.93
8.5849
.71172
5.41295
3.53
12.4609
1.87883
5.94138
4.13
17.0569
2.03224
6.42651
2.94
8.6436
.71464
5.42218
3.54
12.5316
1.88149
5.94979
4.14
17.1396
2.03470
6.43428
1.99
8.7025
1.71756
5.43139
8.89
12.6025
1.88414
5.95819
4.19
17.2225
2.03715
6.44205
2.96
8.7616
1.72047
5.44059
3.56
12.6736
1.88680
5.96657
4.16
17.3056
2.03961
6.44981
2.97
8.8209
1.72337
5.44977
3.57
12.7449
1.88944
5.97495
4.17
17.3889
2.04206
6.45755
2.98
8.8804
1.72627
5.45894
3.58
12.8164
1.89209
5.98331
4.18
17.4724
2.04450
6.46529
2.99
8.9401
1.72916
5.46809
3.59
12.8881
1.89473
5.99166
4.19
17.5561
2.04695
6.47302
3.00
9.0000
1.73205
5.47723
3.60
12.9600
1.89737
6.00000
4JO
17.6400
2.04939
6.48074
3.01
9.0601
1.73494
5.48635
3.61
13.0321
1.90000
0.00833
4.21
17.7241
2.05183
6.48845
3.02
9.1204
1.73781
5.49545
3.62
13.1044
1.90263
6.01664
4.22
17.8084
2.05426
6.49615
3.03
9.1809
1.74069
5.50454
3.63
13.1769
1.90526
6.02495
4.23
17.8929
2.05670
6.50384
3.04
9.2416
1.74356
5.51362
3.64
13.2496
1.90788
6.03324
4.24
17.9776
2.05913
6.51153
8.09
9.3025
1.74642
5.52268
3.69
13.3225
1.91050
6.04152
4J9
18.0625
2.06155
6.51920
3.06
9.3636
1.74929
5.53173
3.66
13.3956
1.91311
6.04979
4.26
18.1476
2.06398
6.52687
3.07
9.4249
1.75214
5.54076
3.67
13.4689
1.91572
6.05805
4.27
18.2329
2.06640
6.53452
3.08
9.4864
1.75499
5.54977
3.68
13.5424
1.91833
6.06630
4.28
18.3184
2.06882
6.54217
3.09
9.5481
1.75784
5.55S78
3.69
13.6161
1.92094
6.07454
4.29
18.4041
2.07123
6.54981
3.10
9.6100
1.76068
5.56776
3.70
13.6900
1.92354
6.08276
4.80
18.4900
2.07364
6.55744
3.11
9.6721
1.70352
5.57674
3.71
13.7641
1.92614
6.09098
4.31
18.5761
2.07605
6.56506
3.12
9.7344
1.76635
5.58570
3.72
13.8384
1.92873
6.09918
4.32
18.6624
2.07846
6.57267
3.13
9.7969
1.76918
5.59464
3.73
13.9129
1.93132
6.10737
4.33
18.7489
2.08087
6.58027
3.14
9.8596
1.77200
5.60357
3.74
13.9876
1.93391
6.11555
4.34
18.8356
2.08327
6.58787
8.19
9.9225
1.77482
5.61249
8.78
14.0625
1.93649
6.12372
4.38
18.9225
2.08567
6.59545
3.16
9.9856
1.77764
5.62139
3.76
14.1376
1.93907
6.13188
4.36
19.0096
2.08806
6.60303
3.17
10.0489
1.78045
5.63028
3.77
14.2129
1.94165
6.14003
4.37
19.0969
2.09045
6.61060
3.18
10.1124
1.78326
5.63915
3.78
14.2884
1.94422
6.14817
4.38
19.1844
2.09284
6.61816
8.19
10.1761
1.78606
5.64801
3.79
14.3641
1.94679
6.15630
4.39
19.2721
2.09523
6.62571
840
10.2400
1.78885
5.65685
3.80
14.4400
1.94936
6.16441
4.40
19.3600
2.09762
6.63325
3.21
10.3041
1.79165
5.66569
3.81
14.5161
1.95192
6.17252
4.41
19.4481
2.10000
6.64078
3.22
10.3684
1.79444
5.67450
3.82
14.5924
1.95448
6.18061
4.42
19.5364
2.10238
6.64831
3.23
10.4329
1.79722
5.68331
3.83
14.6689
1.95704
6.18870
4.43
19.6249
2.10476
6.65582
3.24
10.4976
1.80000
5.69210
3.84
14.7456
1.95959
6.19677
4.44
19.7136
2.10713
6.66333
8.18
10.5625
1.80278
5.70088
3.85
14.8225
1.96214
6.20484
4.49
19.8025
2.10950
6.67083
3.26
10.6276
1.80555
5.70964
3.86
14.8996
1.96469
6.21289
4.46
19.8916
2.11187
6.67832
3.27
10.6929
1.80S31
5.71839
3.87
14.9769
1.96723
6.22093
4.47
19.9809
2.11424
6.68581
3.28
10.7584
1.81108
5.72713
3.&8
15.0544
1.96977
6.22896
4.48
20.0704
2.11660
6.69328
3.29
10.8241
1.81384
5.73585
3.89
15.1321
1.97231
6.23699
4.49
20.1601
2.11896
6.70075
8.80
10.8900
1.81659
5.74456
3.90
15.2100
1.97484
6.24500
4.90
20.2500
2.12132
6.70820
3.31
10.9561
1.81934
5.75326
3.91
15.2881
1.97737
6.25300
4.51
20.3401
2.12368
6.71565
3.32
11.0224
1.82209
5.76194
3.92
15.3664
1.97990
6.26099
4.52
20.4304
2.12603
6.72309
3.33
11.0889
1.82483
5.77062
3.93
15.4449
1.98242
6.26897
4.53
20.5209
2.12838
6.73053
3.34
11.1556
1.82757
5.77927
3.94
15.5236
1.98494
6.27694
4.54
20.6116
2.13073
6.73795
8.38
11.2225
1.83030
5.78792
3.98
15.6025
1.98746
6.28490
4.55
20.7025
2.13307
6.74637
3.36
11.2896
1.83303
5.79655
3.96
15.6816
1.98997
6.29285
4.56
20.7936
2.13542
6.75278
3.37
113569
1.83576
5.80517
3.97
15.7609
1.99249
6.30079
4.57
20.8849
2.13776
6.76018
3.38
11.4244
1.83848
5.81378
3.98
15.8404
1.99499
6.30872
4.58
20.9764
2.14009
6.76757
3.39
11.4921
1.84120
5.82237
3.99
15.9201
1.99750
6.31664
4.59
21.0681
2.14243
6.77495
8.40
11.5600
1.84391
5.83095
4.00
16.0000
2.00000
6.32456
4.60
21.1600
2.14476
6.78233
N
N
VN
ViSv
N
N*
VN
Viw
JV
N*
V5r
viw
334
Appendix A
TABLE V (continued)
N
N*
VN
VlOAT
N
N*
VN
VUM
N
N*
VN
VION
4.60
21.1600
2.14476
6.78233
5JO
27.0400
2.28035
7.21110
5.80
33.6400
2.40832
7.61577
4.61
21.2521
2.14709
6.78970
5.21
27.1441
2.28254
7.21803
5.81
33.7561
2.41039
7.62234
4.62
21.3444
2.14942
6.79706
5.22
27.2484
2.28473
7.22496
5.82
33.8724
2.41247
7.62889
4.63
21.4369
2.15174
6.80441
5.23
27.3529
2.28692
7.23187
5.83
33.9889
2.41454
7.63544
4.64
21.5296
15407
6.81176
5.24
27.4576
2.28910
7.23878
5.84
34.1056
2.41661
7.64199
4.65
21.6225
2.15639
6.81909
5.26
27.5625
2.29129
7.24569
5.85
34.2225
2.41868
7.64853
4.66
21.7156
2.15870
6.82642
5.26
27.6676
2.29347
7.25259
5.86
34.3396
2.42074
7.65506
4.67
21.8089
2.16102
6.83374
5.27
27.7729
2.29565
7.25948
5.87
34.4569
2.42281
7.66159
4.68
21.9024
2.16333
6.84105
5.28
27.8784
2.29783
7.26636
5.88
34.5744
2.42487
7.66812
4.69
21.9961
2.16564
6.84836
5.29
27.9841
2.30000
7.27324
5.89
34.6921
2.42693
7.67463
4.70
22.0900
2.16795
6.85565
5.30
28.0900
2.30217
7.28011
5.90
34.8100
2.42899
7.68115
4.71
22.1841
2.17025
6.86294
5.31
28.1961
2.30434
7.28697
5.91
34.9281
2.43105
7.68765
4.72
22.2784
2.17256
6.87023
5.32
28.3024
2.30651
7.29383
5.92
35.0464
2.43311
7.69415
4.73
22.3729
2.17486
6.87750
5.33
28.4089
2.30868
7.30068
5.93
35.1649
2.43516
7.70065
4.74
22.4676
2.17715
6.88477
5.34
28.5156
2.31084
7.30753
5.94
35.2836
2.43721
7.70714
4.75
22.5625
2.17945
6.89202
5.35
28.6225
2.31301
7.31437
5.95
35.4025
2.43926
7.71362
4.76
22.6576
2.18174
6.89928
5.36
28.7296
2.31517
7.32120
5.96
35.5216
2.44131
7.72010
4.77
?2.7529
2.18403
6.90652
5.37
28.8369
2.31733
7.32803
5.97
35.6409
2.44336
7.72658
4.78
22.8484
2.18632
6.91375
5.38
28.9444
2.31948
7.33485
5.98
35.7604
2.44540
7.73305
4.79
22.9441
2.18861
6.92098
5.39
29.0521
2.32164
7.34166
5.99
35.8801
2.44745
7.73951
4.80
23.0400
2.19089
6.92820
5.40
29.1600
2.32379
7.34847
6.00
36.0000
2.44949
7.74597
4.81
23.1361
2.19317
6.93542
5.41
29.2681
2.32594
7.35527
6.01
36.1201
2.45153
7.75242
4.82
23.2324
2.19545
6.94262
5.42
29.3764
2.32809
7.36206
6.02
36.2404
2.45357
7.75887
4.83
23.3289
2.19773
6.94982
5.43
29.4849
2.33024
7.36885
6.03
36.3609
2.45561
7.76531
4.84
23.4256
2.20000
6.95701
5.44
29.5936
2.33238
7.37564
6.04
36.4816
2.45764
7.77174
4.85
23.5225
2.20227
6.96419
5.45
29.7025
2.33452
7.38241
6.06
36.6025
2.45967
7.77817
4.86
23.6196
2.20454
6.97137
5.46
29.8116
2.33666
7.38918
6.06
36.7236
2.46171
7.78460
4.87
23.7169
2.20681
6.97854
5.47
29.9209
2.33880
7.39594
6.07
36.8449
2.46374
7.79102
4.88
23.8144
2.20907
6.98570
5.48
30.0304
2.34094
7.40270
6.08
36.9664
2.46577
7.79744
4.89
23.9121
2.21133
6.99285
5.49
30.1401
2.34307
7.40945
6.09
37.0881
2.46779
7.80385
4.90
24.0100
2.21359
7.00000
5.50
30.2500
2.34521
7.41620
6.10
37.2100
2.46982
7.81025
4.91
24.1081
2.21585
7.00714
5.51
30.3601
2.34734
7.42294
6.11
37.3321
2.47184
7.81665
4.92
24.2064
2.21811
7.01427
5.52
30.4704
2.34947
7.42967
6.12
37.4544
2.47386
7.82304
4.93
24.3049
2.22036
7.02140
5.53
30.5809
2.35160
7.43640
6.13
37.5769
2.47588
7.82943
4.94
24.4036
2.22261
7.02851
5.54
30.6916
2.35372
7.44312
6.14
37.6996
2.47790
7.83582
4.95
24.5025
2.22486
7.03562
6.55
30.8025
2.35584
7.44983
6.16
37.8225
2.47992
7.84219
4.96
24.6016
2.22711
7.04273
5.56
30.9136
2.35797
7.45654
6.16
37.9456
2.48193
7.84857
4.97
24.7009
2.22935
7.04982
5.57
31.0249
2.36008
7.46324
6.17
38.0689
2.48395
7.85493
4.98
24.8004
2.23159
7.05691
5.58
31.1364
2.36220
7.46994
6.18
38.1924
2.48596
7.86130
4.99
24.9001
2.23383
7.06399
5.59
31.2481
2.36432
7.47663
6.19
38.3161
2.48797
7.86766
5.00
25.0000
2.23607
7.07107
5.60
31.3600
2.36643
7.48331
6.20
38.4400
2.48998
7.87401
5.01
25.1001
2.23830
7.07814
5.61
31.4721
2.36854
7.48999
6.21
38.5641
2.49199
7.88036
5.02
25.2004
2.24054
7.08520
5.62
31.5844
2.37065
7.49667
6.22
38.6884
2.49399
7.88670
5.03
25.3009
2.24277
7.09225
5.63
31.6969
2.37276
7.50333
6.23
38.8129
2.49600
7.89303
5.04
25.4016
2.24499
7.09930
5.64
31.8096
2.37487
7.50999
6.24
38.9376
2.49800
7.89937
5.05
25.5025
2.24722
7.10634
5.65
31.9225
2.37697
7.51665
6.25
39.0625
2.50000
7.90569
5.06
25.6036
2.24944
7.11337
5.66
32.0356
2.37908
7.52330
6.26
39.1876
2.50200
7.91202
5.07
25.7049
2.25167
7.12039
5.67
32.1489
2.38118
7.52994
6.27
39.3129
2.50400
7.91833
5.08
25.8064
2.25389
7.12741
5.68
32.2624
2.38328
7.53658
6.28
39.4384
2.50599
7.92465
5.09
25.9081
2.25610
7.13442
5.69
32.3761
2.38537
7.54321
6.29
39.5641
2.50799
7.93095
6.10
26.0100
2.25832
7.14143
5.70
32.4900
2.38747
7.54983
6.30
39.6900
2.50998
7.93725
5.11
26.1121
2.26053
7.14843
5.71
32.6041
2.38956
7.55645
6.31
39.8161
2.51197
7.94355
5.12
26.2144
2.26274
7.15542
5.72
32.7184
2.39165
7.56307
6.32
39.9424
2.51396
7.94984
5.13
26.3169
2.26495
7.16240
5.73
32.8329
2.39374
7.56968
6.33
40.0689
2.51595
7.95613
5.14
26.4196
2.26716
7.16938
5.74
32.9476
2.39583
7.57628
6.34
40.1956
2.51794
7.96241
5.15
26.5225
2.26936
7.17635
6.75
33.0625
2.39792
7.58288
6.36
40.3225
2.51992
7.96869
5.16
26.6256
2.27156
7.18331
5.76
33.1776
2.40000
7.58947
6.36
40.4496
2.52190
7.97496
5.17
26.7289
2.27376
7.19027
5.77
33.2929
2.40208
7.59605
6.37
40.5769
2.52389
7.98123
5.18
26.8324
2.27596
7.19722
$.78
33.4084
2.40416
7.60263
6.38
40.7044
2.52587
7.98749
5.19
26.9361
2.27816
7.20417
5.79
33.5241
2.40624
7.60920
6.39
40.8321
2.52784
7.99375
5.20
27.0400
2.28035
7.21110
5.60
33.6400
2.40832
7.61577
6.40
40.9600
2.52982
8.00000
N
N*
V*
Vlotf
N
N 9
VN
viSv
N
N*
VN
vTotf
Appendix A
335
TABLE V (continued)
N
N*
VN
view
N
N*
VJV
Vvw
N
JV
VN
VJSfi
6.40
40.9600
2.52982
8.00000
7.00
49.0000
2.64575
8.36660
7.80
57.7600
2.75681
8.71780
6.41
41.0881
2.53180
8.00625
7.01
49.1401
2.64764
8.37257
7.61
57.9121
2.75862
8.72353
6.42
41.2164
2.53377
8.01249
7.02
49.2804
2.64953
8.37854
7.62
58.0644
2.76043
8.72926
6.43
41.3449
2.53574
8.01873
7.03
49.4209
2.65141
8.38451
7.63
58.2169
2.76225
8.73499
6.44
41.4736
2.53772
8.02496
7.04
49.5616
2.65330
8.39047
7.64
58.3696
2.76405
8.74071
4.46
41.6025
2.53969
8.03119
7.08
49.7025
2.65518
8.39643
7.88
58.6225
2.76586
8.74643
6.46
41.7316
2.54165
8.03741
7.06
49.8436
2.65707
8.40238
7.66
58.6756
2.76767
8.75214
6.47
41.8609
2.54362
8.04363
7.07
49.9849
2.65895
8.40833
7.67
58.8289
2.76948
8.75785
6.48
41.9904
2.54558
8.04984
7.08
50.1264
2.66083
8.41427
7.68
58.9824
2.77128
8.76356
6.49
42.1201
2.54755
8.05605
7.09
50,2681
2.66271
8.42021
7.69
59.1361
2.77308
8.76926
.50
42.2500
2.54951
8.06226
7.10
50.4100
2.66458
8.42615
7.70
59.2900
2.77489
8.77496
6.61
42.3801
2.55147
8.06846
7.11
50.5521
2.66646
8.4321)6
7.71
59.4441
2.77669
8.78066
6.52
42.5104
2.55343
8.07465
7.12
50.6944
2.66833
8.43801
7.72
59.5984
2.77849
8.78635
6.53
42.6409
2.55539
8.08084
7.13
50.8369
2.67021
8.44393
7.73
59.7529
2.78029
8.79204
6.54
42.7716
2.55734
8.08703
7.14
50.9796
2.67208
8.44985
7.74
59.9076
2.78209
8.79773
6.88
42.9025
2.55930
8.09321
7.15
51.1225
2.67395
8.45577
7.78
60.0625
2.78388
8.80341
6.56
43.0336
2.56125
8.09938
7.16
51.2656
2.67582
8.46168
7.76
60.2176
2.78568
8.80909
6.57
43.1649
2.56320
8.10555
7.17
51.4089
2.67769
8.46759
7.77
60.3729
2.78747
8.81476
6.58
43.2964
2.56515
8.11172
7.18
51.5524
2.67955
8.47349
7.78
60.5284
2.78927
8.82043
6.59
43.4281
2.56710
8.11788
7.19
51.6961
2.68142
8.47939
7.79
60.6841
2.79106
8.82610
.60
43.5600
2.56905
8.12404
7 JO
51.8400
2.68328
8.48528
7.80
60.8400
2.79285
8.83176
6.61
43.6921
2.57099
8.13019
7.21
51.9841
2.68514
8.49117
7.81
60.9961
2.79464
8.83742
6.62
43.8244
2.57294
8.13634
7.22
52.1284
2.68701
8.49706
7.82
61.1524
2.79643
8.84308
6.63
43.9569
2.57488
8.14248
7.23
52.2729
2.68887
8.50294
7.83
61.3089
2.79821
8.84873
6.64
44.0896
2.57682
8.14862
7.24
52.4176
2.69072
8.50882
7.84
61.4656
2.80000
8.85438
6.M
44.2225
2.57876
8.15475
7J8
52.5625
2.69258
8.51469
7.88
61.6225
2.80179
8.86002
6.66
44.3556
2.58070
8.16088
7.26
52.7076
2.69444
8.52056
7.86
61.7796
2.80357
836566
6.67
44.4889
2.58263
8.16701
7.27
52.8529
2.69629
8.52643
7.87
61.9369
2.80535
8.87130
6.68
44.6224
2.58457
8.17313
7.28
52.9984
2.69815
8.53229
7.88
62.0944
2.80713
8.87694
6.69
44.7561
2.58650
8.17924
7.29
53.1441
2.70000
8.53815
7.89
62.2521
2.80891
8.88257
.70
44.8900
2.58844
8.18535
7.30
53.2900
2.70185
8.54400
7.90
62.4100
2.81069
8.88819
6.71
45.0241
2.59037
8.19146
7.31
53.4361
2.70370
8.54985
7.91
62.5681
2.81247
8.89382
6.72
45.1584
2.59230
8.19756
7.32
53.5824
2.70555
8.55570
7.92
62.7264
2.81425
8.89944
6.73
45.2929
2.59422
8.20366
7.33
53.7289
2.70740
8.56154
7.93
62.8849
2.81603
8.90505
6.74
45.4276
2.59615
8.20975
7.34
53.8756
2.70924
8.56738
7.94
63.0436
2.81780
8.91067
'.76
45.5625
2.59808
8.21584
7.38
54.0225
2.71109
8.57321
7.98
63.2025
2.81957
8.91628
6.76
45.6976
2.60000
8.22192
7.36
54.1696
2.71293
8.57904
7.96
63.3616
2.82135
8.92188
6.77
45.8329
2.60192
8.22800
7.37
54.3169
2.71477
8.58487
7.97
63.5209
2.82312
8.92749
6.78
45.9684
2.60384
8.23408
7.38
54.4644
2.71662
8.59069
7.98
63.6804
2.82489
8.93308
6.79
46.1041
2.60576
8.24015
7.39
54.6121
2.71846
8.59651
7.99
63.8401
2.82666
8.93868
.80
46.2400
2.60768
8.24621
7.43
54.760Q
2.72029
8.60233
8.00
64.0000
2.82843
8.94427
6.81
46.3761
2.60960
8.25227
7.41
54.9081
2.72213
8.60814
8.01
64.1601
2.83019
8.94986
6.82
46.5124
2.61151
8.25833
7.42
55.0564
2.72397
8.61394
8.02
64.3204
2.83196
8.95545
6.83
46.6489
2.61343
8.26438
7.43
55.2049
2.72580
8.61974
8.03
64.4809
2.83373
8.96103
6.84
46.7856
2.61534
8.27043
7.44
55.3536
2.72764
8.62554
8.04
64.6416
2.83549
8.96660
.8ft
46.9225
2.61725
8.27647
7.48
55.5025
2.72947
8.63134
8.08
64.8025
2.83725
8.97218
6.86
47.0596
2.61916
8.28251
7.46
55.6516
2.73130
8.63713
8.06
64.9636
2.83901
8.97775
6.87
47.1969
2.62107
8.28855
7.47
55.8009
2.73313
8.64292
8.07
65.1249
2.84077
8.98332
6.88
47.3344
2.62298
8.29458
Z.48
55.9504
2.73496
8.64870
8.08
65.2864
2.84253
8.98888
6.89
47.4721
2.62488
8.30060
7.49
56.1001
2.73679
8.65448
8.09
65.4481
2.84429
8.99444
8.90
47.6100
2.62679
8.30662
7.80
56.2500
2.73861
8.66025
8.10
65.6100
2.84605
9.00000
6.91
47.7481
2.62869
8.31264
7.51
56.4001
2.74044
8.66603
8.11
65.7721
2.84781
9.00555
6.92
47.8864
2.63059
8.31865
7.52
56.5504
2.74226
8.67179
8.12
65.9344
2.84956
9.01110
6.93
48.0249
2.63249
8.32466
7.53
66.7009
2.74408
8.67756
8.13
66.0969
2.85132
9.01665
6.94
48.1636
2.63439
8.33067
7.54
56.8516
2.74591
8.68332
8.14
66.2596
2.85307
9.02219
8.98
48.3025
2.63629
8.33667
7.88
57.0025
2.74773
8.68907
8.18
66.4225
2.85482
9.02774
6.96
48.4416
2.63818
8.34266
7.56
57.1536
2.74955
8.69483
8.16
66.5856
2.85657
9.03327
6.97
48.5809
2.64008
8.34865
7.67
57.3049
2.75136
8.70057
8,17
66.7489
2.85832
9.03881
6.98
48.7204
2.64197
8.35464
7.58
57.4564
2.75318
8.70632
8.18
66.9124
2.86007
9.04434
6.99
48.8601
2.64386
8.36062
7.59
57.6081
2.75500
8.71206
8.19
67.0761
2.86182
9.04986
7.00
490000
2.64575
8.36660
7.80
57.7600
2.75681
8.71780
8,10
67.2400
2.86356
9.05539
N
N
VN
van
AT
Af
VN
vlitf
N
AT
Vfi
van
336
Appendix A
TABLE V (continued)
N
W
VN
viOAT
N
AT*
VN
VlOAT
N
N*
VS
Viw
8 JO
67.2400
2.86356
9.05539
8.80
77.4400
2.96648
9.33083
9.40
88.3600
3.06594
9.69530
8.21
67.4041
2.86531
9.06091
8.81
77.6161
2.96816
9.38616
9.41
88.5481
3.06757
9.70052
8.22
67.5684
2.86702
9.06642
8.82
77.7924
2.96985
9.39149
9.42
83.7364
3.06920
9.70567
8.23
67.7329
2.8688C
9.07193
8.83
77.9639
2.97153
9.39631
9.43
83.9249
3.07033
9.71082
8.24
67.8976
2.87054
9.07744
8.84
78.1456
2.97321
9.40213
9.44
89.1136
3.07246
9.71597
SJS
68.0625
2.87228
9.08295
8.86
78.3225
2.97489
9.40744
9.45
89.3025
3.07409
9.72111
8.26
68.2276
2.87402
9.08845
8.86
78.4996
2.97658
9.41276
9.46
89.4916
3.07571
9.72625
8.27
68.3929
2.87576
9.09395
8.87
78.6769
2.97826
9.41807
9.47
89.6809
3.07734
9.73139
8.28
68.5584
2.87750
9.09945
8.88
78.8544
2.97993
9.42333
9.48
89.8704
3.07896
9.73653
8.29
68.7241
2.87924
9.10494
8.89
79.0321
2.98161
9.42863
9.49
90.0601
3.08058
9.74166
8.90
68.8900
2.88097
9.11043
8.90
79.2100
2.98329
9.43398
9.60
90.2500
3.08221
9.74679
8.31
69.0561
2.88271
9.11592
8.91
79.3881
2.98496
9.43923
9.51
90.4401
3.08333
9.75192
8.32
69.2224
2.88444
9.12140
8.92
79.5664
2.98664
9.44458
9.52
90.6304
3.03545
9.75705
8.33
69.3889
2.88617
9.12633
8.93
79.7449
2.98831
9.44987
9.53
90.8209
3.08707
9.76217
8.34
69.5556
2.88791
9.13236
8.94
79.9236
2.98998
9.45516
9.54
91.0116
3.08869
9.76729
&*9
69.7225
2.88964
9.13783
8.95
80.1025
2.99166
9.46044
9.55
91.2025
3.09031
9.77241
8.36
69.8896
2.89137
9.14330
8.96
30.2816
2.99333
9.46573
9.56
91.3936
3.09192
9.77753
8.37
70.0569
2.89310
9.14877
8.97
30.4609
2.99500
9.47101
9.67
91.5349
3.09354
9.78264
8.38
70.2244
2.89482
9.15423
8.98
80.6404
2.99666
9.47629
9.58
91.7764
3.09516
9.78775
8.39
70.3921
2.89655
9.15969
8.99
80.8201
2.99833
9.48156
9.59
91.9681
3.09677
9.79285
8.40
70.5600
2.89823
9.16515
9.00
81.0000
3.00000
9.48683
9.60
92.1600
3.09839
9.79796
8.41
70.7281
2.90000
9.17061
9.01
81.1301
3.00167
9.49210
9.61
92.3521
3.10000
9.30306
8.42
70.8964
2.90172
9.17606
9.02
81.3604
3.00333
9.49737
9.62
92.5444
3.10161
9.30316
8.43
71.0649
2.90345
9.18150
9.03
81.5409
3.00500
9.50263
9.63
92.7369
3.10322
9.81326
8.44
71.2336
2.90517
9.18695
9.04
81.7216
3.00666
9.50789
9.64
92.9296
3.10483
9.81835
8.45
71.4025
2.90689
9.19239
9.05
81.9025
3.00832
9.51315
9.65
93.1225
3.10644
9.82344
8.46
71.5716
2.90861
9.19733
9.06
82.0836
3.00998
9.51840
9.66
93.3156
3.10805
9.82353
8.47
71.7409
2.91033
9.20326
9.07
82.2649
3.01164
9.52365
9.67
93.5089
3.10966
9.83362
8.48
71.9104
2.91204
9.20869
9.03
32.4464
3.01330
9.52890
9.68
93.7024
3.11127
9.83370
8.49
72.0801
2.91376
9.21412
9.09
82.6281
3.01496
9.53415
9.69
93.8961
3.11238
9.84378
8.50
72.2500
2.91543
9.21954
9.10
82.3100
3.01662
9.53939
9.70
94.0900
3.11448
9.84886
8.51
72.4201
2.91719
9.22497
9.11
82.9921
3.01828
9.54463
9.71
94.2841
3.11609
9.85393
8.52
72.5904
2.91890
9.23033
9.12
83.1744
3.01093
9.54987
9.72
94.4784
3.11769
9.85901
8.53
72.7609
2.92062
9.23580
9.13
83.3569
3.02159
9.55510
9.73
94.6729
3.11929
9.86408
8.54
72.9316
2.92233
9.24121
9.14
83.5396
3.02324
9.56033
9.74
94.8676
3.12090
9.86914
8.55
73.1025
2.92404
9.24662
9.15
83.7225
3.02490
9.56556
9.76
95.0625
3.12250
9.87421
8.56
73.2736
2.92575
9.25203
9.16
83.9056
3.02655
9.57079
9.76
95.2576
3.12410
9.87927
8.57
73.4449
2.92746
9.25743
9.17
84.0889
302820
9.57601
9.77
95.4529
3.12570
9.88433
8.58
73.6164
2.92916
9.26283
9.18
84.2724
3.02985
9.58123
9.78
95.6484
3.12730
9.88939
8.59
73.7881
2.93087
9.26823
9.19
84.4561
3.03150
9.58645
9.79
95.8441
3.12890
9.89444
8.60
73.9600
2.93258
9.27362
9JO
34.6400
3.03315
9.59166
9.80
96.0400
3.13050
9.89949
8.61
74.1321
2.93423
9.27901
9.21
84.8241
3.03480
9.69687
9.81
96.2361
3.13209
9.90454
8.62
74.3044
2.93593
9.23440
9.22
85.0084
3.03645
9.60208
9.82
96.4324
3.13369
9.90959
8.63
74.4769
2.93769
9.28973
9.23
85.1929
3.03809
9.60729
9.83
96.6289
3.13528
9.91464
8.64
74.6496
2.93939
9.29516
9.24
85.3776
3.03974
9.61249
9.84
96.8256
3.13688
9.91968
8.W
74.8225
2.94109
9.30054
9J5
85.5625
3.04138
9.61769
9.85
97.0225
3.13847
9.92472
8.66
74.9956
2.94279
9.30591
9.26
85.7476
3.04302
9.62289
9.86
97.2196
3.14006
9.92975
8.67
75.1689
2.94449
9.31123
9.27
85.9329
3.04467
9.62808
9.87
97.4169
3.14166
9.93479
8.68
75.3424
2.94618
9.31665
9.28
861184
3.04631
9.63328
9.88
97.6144
3.14325
9.93982
8.69
75.5161
2.94738
9.32202
9.29
86.3041
3.04795
9.63846
9.89
97.8121
3.14484
9.94485
8.70
75.6900
2.94958
9.32738
9.80
86.4900
3.04959
9.64365
9.90
98.0100
3.14643
9.94987
8.71
75.8641
2.95127
9.33274
9.31
86.6761
3.05123
9.64883
9.91
98.2081
3.14802
9.95490
8.72
76.0384
2.95296
9.33809
9.32
86.8624
3.05287
9.65401
9.92
98.4064
3.14960
9.95992
8.73
76.2129
2.95466
9.34345
9.33
87.0489
3.05450
9.65919
9.93
98.6049
3.15119
9.96494
8.74
76.3876
2.95635
9.34830
9.34
87.2356
3.05614
9.66437
9.94
98.8036
3.15278
9.96995
8.75
76.5625
2.95804
9.35414
9.35
87.4225
3.05778
9.66954
9.95
99.0025
3.15436
9.97497
8.76
76.7376
2.95973
9.35949
9.36
87.6096
3.05941
9.67471
9.96
99.2016
3.15595
9.97998
8.77
76.9129
2.96142
9.36433
9.37
87.7969
3.06105
9.67983
9.97
99.4009
3.15753
9.98499
8.78
77.0884
2.96311
9.37017
9.38
87.9344
3.06268
9.68504
9.98
99.6004
3.15911
9.98999
8.79
77.2641
2.96479
9.37550
9.39
88.1721
3.06431
9.69020
9.99
99.8001
3.16070
9.99500
8.80
77.4400
2.96648
9.38033
9.40
88.3600
3.06594
9.69536
0.00
100.000
3.16228
10.0000
N
AP
VN
VION
N
AP
VN
ViiN
N
N*
VN
VtQN
Answers fo
Odd-Numbered
Problems
EXERCISE 1-1. PAGE 11
1. (a) Commutative law of addition.
(c) Associative law of multiplication.
(e) Commutative law of multiplication.
2. (a) 2. (c) 1. (e) 1. (g) -5. (i) -2. (k) -7.
3. (a) -6. (c) -35. (e) -10. (g) 2. (i) 29.
4. (a) -5. (c) 0. (e) -2/3. (g) 36 - 2a. (i) -a6. (k) 3 - x.
5. (a)l. (c)5/17. (e) 1/1.02. (g)-||- x+y. (k) r - 0.1.
EXERCISE 1-2. PAGE 15
1. (a) -3, -2, 0, 4, 5. (c) -4, -2, -1, -1/3, 10. (e) -2, -3/2, 1, x/3, 3.
2. (a) 3 > 1/3. (c) V2 > 1.414. (e) 22/7 > IT.
3. (a) False, (c) True, (e) False.
4. (a) 0. (c) 0. (e) 2. (g) ^6. (i) 0.
5. (a) -1 < x < 1. (c) -a x a. (e) -2 < x < 4.
6. (a) 4 < 5 < 6. (c) -1 < < 1. (e) 1 < V3 < 2.
7. a + 7 10.
EXERCISE 1-3. PAGE 20
1.32. 3. -1. 5.1,000,000. 7. -216. 9.2401. 11. 13.
lu
01
15. 17. 32. 19. 125a. 21. a 7 . 23. a 4 6 4 . 25. a 4 6. 27. a".
29. a 2 6 2 *. 31. a 4 . 33. ~- 35. ^- 37. 5 - 6. 39. 46. 41. 3a - 46.
a 4 a 4
43. -2a - 36. 45. (a - 2c)x* - (a + 2b)xy + (6 - I)?/ 2 . 47. 4a - 56 - 2c.
49. a + (6 + c), a - (-6 - c). 51. a 2 + (c 2 - 6 2 ), a 2 - (6 2 - c 2 ).
53. 2a + (6 - 3c), 2a - (3c - 6). 55. x 2 + (- ?/ 2 - z 2 ), z 2 - (y 2 + z 2 ).
57. -a 3 6 + (a& + 6 2 ), -a 3 6 - (- ab - 6 2 ).
59. 2.r + (- 3y - 4z), 2x - (3y + 4s). 61. 12. 63. 20. 65. 17. 67. 1.
69. -6. 71. 8. 73. 10. 75. 13. 77. 13. 79. |- 81. ~-
7 14
EXERCISE 1-4. PAGE 24
I. 2xy. 3. x*y*. 5. 5a 2 - 66 2 . 7. -4a. 9. z 2 + 2xy.
II. a: 3 + 3^ 2 - 3.r + 2. 13. -4.r//. 15. 5x 2 // 2 . 17. 3a 2 + 26 2 .
19. -2a + 26 - 2c. 21. z 2 - 2a;?/ -f y*. 23. ^ x 3 - ^ x 2 + ~ o^ + y y.
25. -18x?/. 27. -lOx- 3 ?/ 2 . 29. 24a 4 /> 2 c. 31. -Go;* - 4jy. 33.
35. 2x 2 + a-?/ - 3?y 2 . 37. 4.T 4 - Sx*y - 4.r 2 ?/ 2 . 39. x* - 2xy* -f i/ 3 - 41. -2*.
43. 2z?/. 45. -1. 47. 2. 49. -z?/ -h 2x - 3y. 51. x + 5. 53. .-c + y.
55. .? 3 + z 2 y + xy* + 2/ 3 . 57. x 3 - .r?/ + y 2 , 59. .r 2 + xy + y 2 . 61. 2y - xy
63. x 3 + 3x 2 + 3a: + 1 ; x 2 + 2x + 1. 65. 2.r 4 - 9a: + 6. 67. Odd values.
339
340 Appendix B
EXERCISE 1-5. PAGE 30
1. 3(x + 2y). 3. 2(2x + 7). 5. a(3x + 2). 7. -x(a - 2c + x\
9. a(x - 2?/ + 3z). 11. ?/ 2 (5 + 3?/ - a). 13. (.r - 4) (x + 4).
15. (2x - 3) (2* 4- 3). 17. (7x - 11) (7* + 11). 19. (3y - a) (3y 4- a).
21. a 2 z 2 (z - 3a) (x + 3a). 23. (0.1 - 6) (0.1 4- 6).
25. (7xy - I2ab) (7xy 4- 12a6). 27. j(0 + .r 2 ) (6 - a; 2 ).
29. 2(x + 2) (x* - 2x 4- 4). 31. (jy 4- z 2 ) (* 2 ?/ 2 - ay* 2 4- * 4 ).
33. (5p 2 <? 3 4- r 6 ) (25p 4 g fl - 5p 2 ry 3 r 6 4- r 10 ).
35. (3 4- */ 2 ) (x - 2/ 2 ) (* 4 4- *V + ?/ 8 ).
37. 3[3s 2 - (2y 4- 3z)] [9.r 4 4- 3x 2 (2y 4- 3z) 4- (2y 4- 3z) 2 ].
39. (x 4- y 4- w) [(x + 2/) 2 4- (x 4- Z/) (z - w) -f (z - ?^) 2 ].
41. (te - y) (36a: 2 4- fay 4- 2/ 2 ). 43. 3(6 - .r 2 ) 2 . 45. (.r?/ - 9) 2 .
47. ( X 2 _ 5) ( X 2 _ 3). 49. (4. r _ 3) (2 X -f 1). 51. ( x _ 2?/) 2 .
53. (x - 4) (X + 3). 55. (a- 2 4- 1) (* 2 4- 2). 57. (1 4- 15.r 3 ) 2 .
59. 5x(x 4- 4) (x - 2). 61. (2x + 1) (x - 6). 63. (2x - 1) (x + 3\
65. (6.r - 1) (x - 6). 67. (x - 0.6) 2 . 69. (a 4- 3) f.r 4- 2y).
71. (4a; 2 - 5) (2x - 3). 73. (x 4- 3) (2a + y - z). 75. (x 2 - 3) (x + 1).
EXERCISE 1-6. PAGE 33
1. 2. 3. 1. 5. x - y. 7. 3x 3 ?/. 9. 2xy*z. 11. 1. 13. 24. 15. 24,r?/z.
17. SOxtyV. 19- x 2 - 4. 21. (x 2 - 49) (x - 3) (x* - 8) (.r 2 - 4).
EXERCISE 1-7. PAGE 35
1. (a) 12. (c) 4a. (e) l&r// 3 . (g) .r - 1. (i) (a - x) (b + x).
101 . . 3a: 2 ?/ , ().r?y 3 1 ... 3>c - 7 ,, 3>c 4-2
2 ' (a) 347' (C) ~IT' (e) i^T5* (g) ^i' W x4-2* W 2*+8*
x 2 - (y - 3) 2 _ , , ^ _ . _ , _ 3a+b, _ x
WT^&rr^r- (0)^-3,. (q)i. (s)^ 6 ^r+,-2^
EXERCISE 1-8. PAGE 38
,17 94 _ 57 _ -2 _ x' - 3.r 2 - 6 ,, x 2 + 3x - 5
L 24" 3 --2l' 5 --20' 7 'x^T- 9 - x3-l * ?
x 2 + x + 4 3* 2 + 12f + 5 1+* 2
*** /. -I \ / . t \ o J.W. / __ . rtx / _ . o\ /_. i ^\ M. I
(x + 1) (x - I) 2 *"' (x + 2) (a; + 3) (x + 4) 1 - ,
iq - x(x + 5y) 21 2(a 2 - ab 4- 6 2 ) q 2j?4-?/4-3 2 - ^ 2 - xy 4- y 2
(*+y)(-y) >- a 2 -6 2 " ** ^?y ' * x 2 -y 2
EXERCISE 1-9. PAGE 41
. 54 n x -5.a n 7a 3 6 4 ..
' 27 ' 2(x + 5)
(g - 1) (x + 2) (9x 2 + (to + 4) (2.t - 5) (4x + 1) (x - 6)
(x - 7) (* - 2) ' 1& " 9x 2 - 4 ' "' (fa + 1) (* - 5) '
Appendix B 341
- 3(4s + 5) (2s + 3) 28 70 32/
5) -- 2L V 23 '59* 25 '5T
33 .i|. 35.1^. 37.2i. 39.2. 41.
x + 5 5z 2 - 1 zy
EXERCISE 1-10. PAGE 47
1. x = -5/2. 3. x = -3/7. 5. x = -2. 7. ^ 2 - 9. -11. 11.7. 13.10.
15
tK -. 4 -23s tft x - 1 01 60; -3
15. -2. 17. y = ^ -- 19- y = ~y~~ ' 21. y = 2
23. y = ~ 2 ' g ,' f 17 25. -5. 27.7. 29.1/4. 31.3. 33.-^- 35. ~
4 J 4
007 4-^8
37. ^jp 39. ~ 41. 4. 43. 45. 2. 47. -- 49. 40, 58. 51. 6, 7.
53. 921,600 sq ft. 55. 170 adults, 330 children. 57. 20, 40, 120.
EXERCISE 2-1. PAGE 52
2. (a) 5. (c) 4. (e) V34. (g) \/29.
3. (a) 5. (c) V2. (e) 7. (g) 4. (i) V^T^. (k) 2.
EXERCISE 2-2. PAGE 56
I. The area of a circle is a function of the radius of the circle, A = Trr 2 .
3. The area of a trapezoid is a function of its altitude and bases, A =-(bi +62).
5. The volume of a cylinder is a function of its height and the radius of its base,
V = irr*h.
7. The annual premium of a life insurance policy is a function of the applicant's age
and physical condition, of the type of policy, of the company's rate policy, etc.
No formula can be written.
9. -3, -5, 3, -2, 2 V2 -3, -3, 3/4, 2y - 3, | - 3,^3'
II. - ~ , 288, Cy 2 + 7. 13. 0. 15. 14. 17. p 2 + q* + r 3 . 19. 0. 21. 30.
oZ
23. 5/7. 25. 0. 27. A = HT*, C = 2xr, A = , C = 2
29. S = ^/3(57r F 2 . 31. all x. 33. all x. 35. all x. 37. all x. 39. all a.
41. all x. 43. all x. 45. s * 0, 1. 47. \x \ 3. 49. x = 0. 51. all s.
53. 3 5* 0, -2. 55. all x. 57. a: -1. 59. -2 ^ x g 2.
EXERCISE 2-3. PAGE 57
1. 3,2.3, | x | . 3. 0,0.5. .5. 1,0,1,0,1.
342 Appendix B
EXERCISE 2-4. PAGE 59
1. y = y x. 3. 567. 5. 5. 7. 10/3. 9. -2. 13. 9/4, 27/8.
15. 1000/1, 100/1. 17. 327C. 19. 99.5 Ib, 95.2 Ib. 21. 0.0324 in.
EXERCISE 3-1. PAGE 67
1. (a) (1, 0). (c) (0, 1). (e) (1, 0).
2. (a) (0.54,0.84). (c) (-0.99,0.14). (e) (-0.65, -0.76).
3. (a) V5/2. (c) 1. (e) -1/2. (g) -2. (i) - V5/2.
7. sin t cos t tan t cot t sec t esc t.
(a) V3/2 I/ V3 V% 2/V3 2.
(c) 5/13 12/13 5/12 12/5 13/5.
(e) A/3/2 - 1/2 V V V3 2/ V3.
(g) 1/V5 2/ \/5 1/2 V5/2 V5.
(i) 4/5 3/4 4/3 5/4 -5/3.
EXERCISE 3-3. PAGE 74
1. (a) 1. (c) 0.58. (e) 0.86.
2. (a) 0.9927. (c) 24.52. (e) -1.500. (g) 0.2571. (i) 5.798. (k) 1.011.
EXERCISE 3-4. PAGE 80
1. 7T/3. 3. 7T/6- 5. 27T/3. 7. 7T/15. 9. 47T/3. 11. 27T/5. 13. 437T/36.
15. 1077T/60. 17. 77T/20. 19. 0.8090. 21. 1.4358. 23. 3.2107. 25. 1.6323.
27. 45. 29. 270. 31. 15. 33. 630. 35. 27. 37. 47023'54". 39. 4343'.
57. 147T/3, 1.25 radians. 59. 14.74 in.
61. (a) 4 radians, (b) 16/9 radians, (c) 0.04 radian.
EXERCISE 3-5. PAGE 85
1. 0.5925. 3. 1.092. 5. 0.7412. 7. -1.453. 9. -0.2462. 11. 0.2504.
13. 1.181. 15. 9.010. 17. 0.6817. 19. 0.8437. 21. 0.9831. 23. 0.5154.
25. 0.2930. 27. -9.462. 29. -1.059. 31. 17310', 35310'.
33. 4230', 22230'. 35. 310', 18310'. 37. 3520', 14440'.
39. 5630', 23630'. 41. 26450', 27510'. 43. 8310', 26310'.
45. 5520', 23520'. 47. 427', 2227'. 49. 189', 34151'. 51. 3135 ; , 32825'.
53. 674', 2474 r . 55. 730', 18730'. 57. 9736', 26224'. 59. 9311' 7 273 !!'.
61. 0.8016. 63. 3.079. 65. -100.00. 67. 0.3459. 69. -0.7073. 71. 1.214.
73. 0.220, 6.060. 75. 1.120, 4.260. 77. 0.755, 3.895. 79. 1.158, 5.122.
81. 1.143, 1.997. 83. 0.574, 5.706.
EXERCISE 4-1. PAGE 96
64 1
1. 3y. 3.^. 5.100. 7.81. 9. ~ 11. ay". 13.1.
Appendix 8
343
21. 4(5"*). 23.
33. (5292) ". 35. s
-...- m _ ^
2 - x x 2 (l + a; 2 )
49. 2(2 - x 2 ) 3 ' 2 .
EXERCISE 4-2. PAGE 100
1. 54. 3. 3/2. 5. 6,652,800. 7. 5/24. 9. n(n - 1). 11. (n + l)n.
13.
15.
71 ~T~ t
19. X 7 _
- 276*.
-f 80x~ 1/2
35a; 3 -
-f- 7x -
80ar 3 / 2 -f-
-f
21. 8a<> -
23. x 612 +
7*8
25. ^--4-+6 -4+
i/4 ?/2 X 4 ^8
27. x 12 - Qx l y 2 + 15z 8 2/ 4 - 2
29. x 2 +y 2 + z 2 + 2xy + 2?/2: + 2x0.
31. z 8 -f 4x 7 + 10x6 -f 16z 5 + 19x 4 + 16x 3 4- 10x 2 + 4* + 1. M- -8x7.
35. 7920a 8 6 4 . 37, -14x 2 . 39. 2 11 3 5 5 7 13x 33 ?/ 8 . 41. 924x 3 i/ 3 .
EXERCISE 5-1. PAGE 104
1. Iog 2 8=3. 3. logs 81 = 4. 5. logio 1000 - 3.
7. Iog 256 2 = I - 9. Iog 100 10 - 0.5. 11. y = Iog 10 x. 13. 8 2 = 64.
15. 2- = ~ 17. 73 = 343. 19. 10 4 = 10,000. 21. 4 3 / 2 = 8. 23. 6. 25. 2.
27. 3. 29. 1/10. 31. 5/2. 33. No solution. 35. Iog 6
37.
39. log* (u - \/u 2 - a 2 ).
41. (a) log* TT + 1/2 Iog 6 1 - 1/2 log b g. (b) 2 log fe 7 T -f Iog 6 g - 2 log* TT.
EXERCISE 5-2. PAGE 107
1. 1. 3. 4. 5. -1. 7. -3. 9. -1. 11. 5. 13. -1. 15. -1. 17. 7314.
19. 7.314. 21. 7314000. 23. 0.007314.
EXERCISE 5-3. PAGE 110
1. 1.5441. 3. 2.0212. 5. 7.7931-10. 7. 4.3636. 9. 9.5490-10. 11. 8.8215-10.
13. 9.9279-10. 15. 0.4536. 17. 7.8452-10. 19. 9.8908-10. 21. 0.4972.
23. 8.9439-10. 25. 9.9824-10. 27. 9.9476-10. 29. 9.1306-10. 31. 46.4.
33. 0.262. 35. 504. 37. 0.0000000000276. 39. 69.2. 41. 292.3.
43. 5,454,000,000. 45. 0.06114. 47. 4.554. 49. 0.00001072. 51. 0.6021.
53. 0.4266. 55. 1.585. 57. 3.728. 59. 2.5023. 61. 1.6297.
344 Appendix B
EXERCISE 5-4. PAGE 112
1. 8.540. 3. 0.04292. 5. 3.183. 7. 0.0008416. 9. 0.1104. 11. 54.61.
13. 48.91. 15. 11,670. 17. 0.1795. 19. 0.02950. 21. 20.56. 23. 538,100.
25.1.708. 27. -1.021. 29.1.249. 31.0.4343,0.2171,9.5657-10,23.1,22.46.
33. 127,900,000 sq ft. 35. 12.62 ft. 37. 6,070,000 sq ft. 39. 16.5 amp.
41. $1,074.00.
EXERCISE 5-5. PAGE 114
1. 2.3026. 3. 1.4429. 5. 0.8735. 7. 6.0001. 9. 1.5373. 11. 2.0794.
13. 1.4307. 15. 0.8228.
EXERCISE 6-1. PAGE 119
1. B = 57, 6 = 18, c = 22. 3. A = 1S50', a = 21, c = 66.
5. A = 27!', B = 6259', c = 7.012. 7. A = 2244', 6 = 10.30, c = 11.17.
9. B = 5339', 6 = 13.40, c = 16.64. 11. B = 4643', a = 73.66, c = 107.5.
13. A = 858', B = 812', c = 793.0. 15. A = 493', a = 2.663, c = 3.528.
17. h = 29 ft, Z = 33 ft. 19. 113 ft. 21. 227 ft. 23. 1334'. 25. 4 ft.
EXERCISE 6-2. PAGE 124
1. 6326', 11,000ft. 3. 60. 5. N1124'W, 74 nautical miles. 7. 400ft.
9. 80 ft, 173 ft.
EXERCISE 6-3. PAGE 131
1. (a) 5. (c) 2 x/5. (e) y/2. (g) 4.
2. (a) 5 [3/5, 4/5]. ( e ) 2 V
(g) 4 [0, 1].
3. (a) 5, 538'. (c) 5, 1438'.
4. (a) VS, 6326'. (c) V73, 15927'.
5. 12, 0. 7. 22 knots, 20 knots. 9. 60 Ib. 11. 49 Ib, 25045 / .
EXERCISE 6-4. PAGE 133
I. 9.8733-10. 3. 8.8059-10. 5. 0.7391. 7. 9.3661-10. 9. 9.9427-10.
II. 9.5906-10. 13. 0.0030. 15. 0.1004. 17. 120', 18120'. 19. 830', 17130'.
21. 18, 198. 23. 5416 / , 23416'. 25. 64 2 r , 29558'. 27. 3011', 210 11'.
29. 5340', 23340'. 31. 2724', 15236'. 33. 127', 181 27'. 35. 6547', 29413'.
EXERCISE 6-5. PAGE 134
1. B = 59, 6 = 60, c = 117. 3. B = 6440', 6 = 133.9, c = 148.2.
5. A = 7115', b = 2.01, a = 5.94. 7. A = 6251', b = 18.53, c = 40.61.
9. 020'. 11. 671 Ib, 20040'. 13. 607 mph, N3642'W.
15. 1815 Ib, 1962 Ib. 17. 49.19 in. 19. 16,900 Ib.
Appendix 8
EXERCISE 7-1. PAGE 138
12-VS- 7. 2 -
345
-. .
11. -3. 13. J (V5 -2 V2), g (1 + 2 V). 15. 4/5, -3/5.
17. 33/56, 63/16.
EXERCISE 7-2. PAGE 142
26
239
169'
, (0-4/5, (g), OOy-
13. cos 30. 15. tan 60. 17. sin 2 20. 19. cos 20. 21. tan 20. 23. cos 9.
11. (a) -3/5, (b) -^ f/50+5x/I6, (c) 3/4, (d) 4/3,
25. cot ? - 27. tan* ^ -
EXERCISE 7-3. PAGE 144
1. ^ [sin 70 - sin 0]. 3. sin 100 + sin 20. 5. ^ [cos 60 -f cos 20].
2t
7. _ 1 [ C o S 48 - cos 8]. 9. - \ [cos 60 - cos 40]. 11. 2 sin ^ cos |.
^ 2i & 2i
Q/a 0/3
13. 2 sin ^ cos ^ 15. - 2 sin 50 sin 30. 17. 2 sin 3230' cos 730'.
2* t
19. 2 sin 43 cos 3.
EXERCISE 8-1. PAGE 154
1. 2?r, 3, 0. 3. 27r, 1/2, 0. 5. Sir/3, 1/3, 0. 7. 5?r/2, , 0.
9. 8 radians, 1, 0. 11. 27T/5, 3, 0. 13. ir/6, , 0. 15. Tr/4, , 0.
4 7
17. 1 radian, , radians. 19. 2 radians, CD, radians. 21. 2ir, 1, 0.
7T 7T
23. TT, 3, ^ radian.
1. x =
EXERCISE 8-2. PAGE 161
3. x = -=- 5. x = -12T/ - 22. 7. Tr/6. 9. ir/6.
2ir
2 m
11. T/2. 13. 7T/3. 15. 7T/6. 17. -7T/3. 19. y 21. -2427'.
5\/6
23. 12/5. 25. 5/13. 27. 3/4. 29. 0.3919. 31.
33. 3/4. 35. -3/5.
47. u. 49. u. 51. 1/u.
37. -7T/2. 39. -w. 41. u. 43. w. 45.
53.
1 -
~ -ir T x j. IT - */er ~
= -==-, 2/ J?H 15. Inconsistent. 17. x = , y = 7/5, 2; =
75 O o o
= y = J ^ = 2L * = 10 / 7 > */ = 25 /7, * = -1/7.
346 Appendix 8
EXERCISE 9-1. PAGE 168
1.x- 10/7, y = -6/7. 3. x = -9/7, y = 15/7. 5. x = -2, y = 1.
18 -1 -47 1 _ -10
to
13.
23. x = 16/11, y = -21/11, z = -9/2. 25. x = 9/5, y = -1/5, z = 1.
27. z = 10/7, y = 4, z = 20/3. 29. x = 0, y = 5/7. 31. Inconsistent.
33. Inconsistent. 35. Consistent and dependent.
37. Consistent and independent. 39. Inconsistent.
41. Inconsistent unless c = 1. Consistent and dependent if c = 1.
EXERCISE 9-2. PAGE T72
1. (1/4, 0), (0, -1/3). 3. (4, 0), (0, -4). 5. (0, 0). 7. (-4/3, 0), (0, 4).
9. (5/3, 0), (0, -5). 11. x = 6/5, y = 4/5. 13. x = 23/7, y = 22/7.
15. x = 1, y = 5/8. 17. x = - j |,y=|.
EXERCISE 10-1. PAGE 180
1.5. 3.0. 5.0. 7.7. 9. -11. 11.22. 13. x = 4.1, y = 0.3.
15. x = 25/13, y = -5/13. 17. (y, o), (0, -17).
19. (-3, -1), (2,4), (6, -3).
EXERCISE 10-2. PAGE 185
1. x = 2, y = 3, z = -2. 3. x = 1, y = -2, 2 = 2/3.
5. z = 2, y = 3, z = 2. 7. = 80, y = 3z. 9. No nontrivial solution.
EXERCISE 10-3. PAGE 187
1. 0. 3. 0. 5. -110. 7. 308. 9. 2184.
EXERCISE 11-1. PAGE 193
I. + 4t, - 4i. 3. - 3i, + 3f. 5. - 6|a|t, + 6|fl|i.
7. 3 \/2 + 3 V2i, 3 \/2 -3 \/2i. 9. 1 + 4 V&, 1 - 4 Vf.
II. \/15 +8|a|Va6i, VT5 -8|a|Va6*'- 13. -i. 15. -t. 17. -i.
19. f. 21. 1. 23. 0. 25. 0. 27. & = 3/2, y = 1/3. 29. z = 6, y = -5.
31. x = 1, y = -4. 33. x = 2, y = 5/2. 35. a? = 7/3, y = 4/3. 37. 7 + 3i.
39. | + ^ t. 41. -1 + K. 43. 25 + Oi. 45. -1 + f. 47. 1 + Oi.
Appendix B 347
49. -3 + (\/2 + \/3)i. 51. 6 + Of. 53. 13 + lit. 55. -8 + 4t.
57. 11 - 3f. 59. 27 + 24t. 61. 28 + IGt. 63. 5 - 2 V + (2 + 5 <>/)
65. -2 - 2(K. 67. ~ + \ i. 69. g - ~ t. 71. 33 - 22t. 73. ~ + ^ i.
75 1 , , 77 625 1019.
75 ' ~ 25 + 25 *' 77 ' 1233 + 3233 * '
EXERCISE 11-2. PAGE 197
13. 3 - 2i. 15. -1 - 9f. 17. 2 + 3f. 19. 6 - 4f. 21. + 3f.
23. 5 + (2 - V5)*. 25. V2(cos 15 + i sin 45). 27. 3(cos 270 + t sin 270).
29. ~^(('os 5444' + t sin 5444'). 31. 13(cos 6723' + f sin 6723').
33. 3 (cos + t sin 0). 35. - (cos 32012' + f sin 32012 / ).
bl
EXERCISE 11-3. PAGE 202
2i. 5 , a
,
A ^
7. 1 + i. 9. -1. 11. -1/2 - ^ i. 13. A + A i. 15. 1+ 1 1.
QSO
17. -2o. 19. | (0.6065 + 0.7951 t). 21. -1.
23. 2[cos (9 + A- 90) + f sin (9 + k 90)], k = 0, 1, 2, 3.
25. 2[cos (36 + k 72) + i (36 + k 72)], fc = 0, 1, 2, 3, 4. 27. 1, t, -1, -t.
29. cos(SO + k 120) + f sin (80 + /<* 120), fc = 0, 1, 2.
ot 14 + 18t 00 828 + 154i ,
31. - - - amperes. 33. - =-=^ - ohms.
EXERCISE 12-1. PAGE 206
1. 0, -7. 3. -1/2, -3/4. 5. 3, -2. 7. 3, 3, -3.
9. sin = 0, 1; 0, 90, 180. 11. sin = 1, -2; 90.
13. sec 0=2, -8; 00, 300, 9711', 26249'.
15. cot = -3, 2; 16134', 34r34', 2(>34', 20634 / .
17. cot 0=7, -17, 88', 1888', 17G38', 35638'.
19. sin = 0, 2, - | ; 0, 180.
EXERCISE 12-2. PAGE 209
i in o Q in Q K -1 =fc V7i - -1 =fc \/21
I. 10, 2. 3. 10, 3. 5. - ^ - 7. - TT -
9. tan = 1 \/2; = 6730', 24730', 15730',
II. sec = 1, -3; 0, 10928', 25032'.
348 Appendix B
13. esc = == 4 ; 36 Q 23', 14337', 23730', 30230 / .
15. cos = 0.4142, -2.414, 6532', 29428'. 17. (x - 3) 2 + 4(y + 2) = 4.
19. (x - 5) 2 + 4(y - 5)2 = 16. 21. 4(s - 2) 2 + 9(i/ - I) 2 = 36.
23. 4(s + 4) 2 - Q(y - 2) 2 = -36. 25. 3(s + 10/3) 2 - (y + I) 2 = 64/3.
27. V2[(z-4) 2 +9/2]. 29. * 31. [2((x + 7) 2 - 32)]~" 2 .
V (3 3) 2 16
33. [9((o; + 4/3) 2 + I)]' 1 / 3 .
EXERCISE 12-3. PAGE 212
I. 1, -7. 3. -1 >/6. 5. -1, -1. 7. """ 3 ^ 9. 5/2, -1/2.
II. -1, - 11/8. 13. -1, -15/7.
15. tan = 1, -5/2, 45, 225, lllW, 29148'.
17. sin 6 = 3.7913, -0.7913; 23218 ; , 30742'.
19. cos = - r - ; no values of 0.
o
21. sec = 0.414, -2.414; 11428 ; , 24532'.
23. esc = -0.6972, -4.3028; 19326', 34634 / .
EXERCISE 12-4. PAGE 213
1. 18. 3. -4. 5. 8 4 x/5. 7. 5. 9. 13 ~
EXERCISE 12-5. PAGE 215
1. d= A/3, 2i. 3. i, it A/6. 5. 2, db3. 7. 5
S
9. -4 =fc \/15, 3 2 A/2.
EXERCISE 12-6. PAGE 216
1. Conjugate imaginary. 3. Real, unequal, irrational.
5. Real, unequal, irrational. 7. Real, unequal, rational.
9. Real, equal, rational. 11. Conjugate imaginary.
13. Real, unequal, irrational. 15. Conjugate imaginary.
EXERCISE 12-7. PAGE 218
1. -2, -1. 3.0,2. 5.3/2,6/5. 7. 6/5, -1/5.
2 J7 _ 2 _
17. s 2 + 1 = 0. 19. z 2 - (\^ - \/5)x =0. 21. x* - 2 A/5 z + 8 = 0.
Appendix B 349
EXERCISE 12-8. PAGE 224
1. Circle. 3. Intersecting lines. 5. Hyperbola. 7. Parabola.
9. Intersecting lines. 11. Hyperbola. 13. Intersecting lines.
15. Intersecting lines. 17. Intersecting lines. 19. Parallel lines.
EXERCISE 12-9. PAGE 227
1. (3, 3), (-3/2, 3/4). 3. (3, 4), (-4, -3). 5. (2, 4), (-3, 9).
7. (3, 2), (-1, -6). 9. (2, 3). 11. (2 V, 1).
13. No solution. 15. (4, 2).
EXERCISE 12-10. PAGE 231
1. (3, 3), (-3/2, 3/4). 3. (3, 4), (-4, -3). 5. (4, -3), (-2, 6).
7. (4, 0), (-5, 3). 9. (4, 6), (-3, -1). 11. (3, 1) (3 VI6i, -10).
13. ( "^ i} 3^85) . , 5 . (2) 4) . . (1> 2)> (2> 1} .
19. (4, 1), (-4, -1), (14, -4), (-14,4).
21. (4, 2), (-4, -2), (V6, -2 V6,), (- V6, 2 VG). 23. (5, 5), (-5, -5).
25. (3, 4) (4, 3), (-3, -4), (-4, -3).
27. (2 + i V2, 2 - i V), (-3 + i \/7, -3 - i \/7), (2 - i V2, 2 + i v^),
(-3-tV7, -3+iV7).
/31 - 3 V93 31+3 Vm\ /31 + 3 x/93 31-3 \/93\
^' V 31 ' 31 )'\ 31 ' 31 / *
EXERCISE 12-11. PAGE 233
1. 6. 3. 4/3. 5. 2.292. 7. 1.682. 9. 1. 11. 1.836. 13. 2.718. 15. 8.547.
17. 2.944. 19. 49.3. 21. 10, 0.1. 23. 0.7S74. 25. -0.44.
27> log(l-ac)-log6-logc t ^ ^- 3h06M4>
log c
EXERCISE 13-1. PAGE 239
I. Q : x - 7, R : 0. 3. Q : x - 1, R : 2. 5. Q : x* - 3x 2 + 6x - 24, R : 78.
7. Q : 2*3 + 3z 2 + 4, 72 : 0. 9. Q : x 2 + 5x + 8, R : 11.
II. Q : x 2 + 2* - 15, R : 0. 13. Q : x* + 3s - 6, R : 0.
15. Q : a;*- 1 + x n ~*y + -f r* 1 , # : 0. 17. 52, 2. 19. 24, -36.
EXERCISE 13-2. PAGE 241
1. 2x* - 2x - 3. 3. Not a factor. 5. x 2 -f 2z - 5.
7. cc' + 2xV + 4a*V -f 8j: 4 2/ + 16x 3 2/ -f 32x^i<> + 64.ri/ + 128y".
9. Not a factor. 11. 5x* + 2a6(5 - 3c* 2 )x - 12a 3 6 4 . 13. Not a factor.
15. 12a 3 - 22* 2 - 34o: + 60. 17. 24s 3 - 9(te 2 + 39z + 45.
350 Appendix B
EXERCISE 13-3. PAGE 244
1. 1 \/3i, -3. 3. 1 =fc i, 3. 5. 2 -\/2 *', 1, 1, 2.
7. x* - 4x* + 9x - 10 = 0.
EXERCISE 13-4. PAGE 247
1. -1/3, 1/2, 5/3. 9. -1. 11. 2, 2, -2, -2. 13. 2/3, " 1 t '
&
15. 1/2, t. 17. 3/5, 1 t. 19. 1/2, V-
EXERCISE 14-1. PAGE 253
I. x < 3. 3. a; > -5. 5. x < 4. 7. a; < -1/2. 9. x < -1.
II. -i<z<i. 13. -|<*<i. 15. -i<*<5. 17. -13 < x < 13.
& & O o & A
19. -4 z 4. 21. -1 < x < 1/3. 23. x < -5/3, s > 2.
25. No values of x. 27. x < -3, -2 < x < -1. 29. 1 < a; < 2, x > 3.
81. | | 2 5. 33. x < -2, s > 0. 35. x < -1/2.
EXERCISE 15-1. PAGE 260
1.1,2,4,7,11. 3.1,2,3,5,11,35. 7. 2, 5, 11, 23, 47; 2, 7, 18, 41, 88.
EXERCISE 15-2. PAGE 263
1. 17, 20. 3. 18, 25. 5. Not an arithmetic progression. 7. 46 30, 5b 40.
9. 4fl ~ 26 , 5a ~ 36 . ll. J 26 = 78, S 26 = 1053. 13. ho = 10, Sio = 55.
& &
15. /7s = 149, S 75 = 5625. 17. / 2 o = 5.9, S 2Q = 61. 19. i 100 = 100, Sioo=5050.
40 fi9
21. /io = 100, S 10 = 550. 23. a = - ^ , Z 45 = ~
o
25. o = 1, n = 13 or a = -1/2, n = 16. 27. / = 9, n = 9. 29. n 2 .
31. d = -^5 . 33. 33j i 6 . 35. 4 - ^~ 1 . 37. $37.75. 39. 282.
11 tt\K ~T~ ly
EXERCISE 15-3. PAGE 264
1. 1/15, 1/19. 3. 1/20, 1/25. 5. 4, 16/5. 7. 1/47. 9. ^, | , y , |.
n 120 60 120 ,
IT" ' T ' "IT '
EXERCISE 15-4. PAGE 267
I. 128, 512. 3. 256, 1024. 5. 8, 16/3. 7. ^j- , ^j" 9. -1,1.
JL& JL&
II. I,. = |J , Si, = 9[1 - (2/3)'*]. 13. l jol = 10-", S 101 =
Ifcfc =!,& ={. 17. 3, -14. 19.i^. 21. 6, 15/2.
16 '"' 16
Appendix B 351
23. 10; 100; 1000; 10,000; 100,000.
27. As n increases, the sum approaches 3 as a limit. 29.
~~ X) JL * X
EXERCISE 15-5. PAGE 270
1.64/65. 3. 9 + 2 5 ' 16 ' 7 ' 3 ' 9 - 5 /8. 11.2/11.
13. 36, 36 [1 - (1/3) 20 ], 36(l/3) 20 . 15. 12 ft, 12 [1 - (3/4) 10 ] feet. 17. 1/9.
19. 10/11. 21. 1/6. 23.3. 25.
EXERCISE 15-6. PAGE 772
-. * ,o * A * . # 2 s 3 ^ - a? , 3z 2 5z 3
1.1 -+-. i.n.--_+_. 8 - 1 -2 + -T-lG'
7 .i_4+*!_y!. 9 .J__f + 6_i?f. 11.1.2190. 13.1.3684.
X X 2 X 3 X* X 3 X* X & X 6
15. 0.8508. 17. 0.9415. 19. 1.0149.
EXERCISE 17-1. PAGE 279
1. 72. 3. 216. 5. 9999. 7. 20,160; 7560.
EXERCISE 17-2. PAGE 282
1. 20; 210; 95,040; 143,640; 970,200. 3. (a) 720, (b) 48, (c) 480. 5. 125.
7. 8 C a 8 4- sCitfb + 8 C 2 a 6 6 2 + sC 3 a 5 6 3 +
Ca/>7 C6 8 . f
9.9,979,200. 11.66. 13.63. 15-
TJ jy
EXERCISE 17-3. PAGE 288
1. 2/5, $24.00. 3. 1/3, 2/3. 5. $7.29. 7. $2.42. 9. 0.553. 11. 16/63,
15 A 215.
15. ,
EXERCISE 18-1. PAGE 295
1. C = 9858', b = 14.55, c = 20.46. 3. A = S927', a = 1169, b = 1079.
5. B = 531', 6 = 1.051, c = 7.513. 7. A = 173', B = 10026', 6 = 71.18.
9. B = 2427 ; , C = 10120 ; , c = 1193. 11. No solution. 13. 615.3 ft.
15. 449 ft. 17. 12.6 in. 19. 3158 Ib,
EXERCISE 18-2. PAGE 298
I. c = 270, B = 5130', A = OO^O 7 . 3. a = 290, B = 1150', C = 101.
5. a = 100, B = 10, C = 20. 7. 4820'. 9. No solution.
II. A = 5420 / , B = 5940', C = 66. 13. 3257 ft. 15. 700 ft,
17. c = 20, A = 63, 5 = 73, C = 44.
352 Appendix 8
EXERCISE 18-3. PAGE 301
1. A = 50, B = 70, c = 56. 3. A = 61, C = 2350', 6 = 262.
5. B = 3111', C = 70, a = 79.25. 7. A = 47, J5 = 58, C = 75.
9. A = 4220', B = 5730 / , C = 8010'. 11. 4 in., 5 in. 13. 11 in.
15. 55 30', 5950', 6440'.
EXERCISE 18-4. PAGE 304
1. 88.41. 3. 243,900. 5. 9285. 7. 90. 11. 54 ft. 15. 45 ft.
17. 38 sq ft, 12 sq ft.
Index
Abscissa, 50
Absolute value, 14, 57
Absolute value function, 57
Addition, 2
of algebraic expressions, 21
of fractions, 36
fundamental laws for, 2
of real numbers, 2
Addition formulas, 135
Algebraic expressions, 17
addition and subtraction of, 21
division of, 23
multiplication of, 22
symbols of grouping of, 18
transposing terms of, 43
Amplitude of a trigonometric
function, 149, 152
Angle
coterminal, 76
definition of, 75
degree measure of, 77
of depression, 120
of elevation, 120
initial side of, 75
measurement of, 76
negative of, 75
positive of, 75
radian measure of, 77
standard position of, 75
terminal side of, 75
trigonometric functions of, 81, 82
vertex of, 75
Antilogarithms, 109
Arc of a circle, 78
Area
of a sector of a circle, 78
of a triangle, 302
Arithmetic means, 262
Associative law
for addition, 2
for multiplication, 2
Axes, coordinate, 49
Axis of symmetry of a parabola, 219
Base
change of logarithmic, 113
exponent and, 16, 86
of logarithms, 101, 113
Bearing in navigation and
surveying, 121
Binomial, 17
Binomial series, 271
Binomial theorem, 97
combinations applied to, 282
general term of, 99
proof of, for positive integral
exponents, 275
Braces, 18
Brackets, 18
Cartesian coordinates, 50
Characteristic of a logarithm, 106
Characteristic of the general
quadratic equation, 222
Circular system, 77
Classification of functions, 61
Coefficient, 17, 18
Combinations, 278, 281
and the binomial coefficients, 282
Common difference, 260
Common logarithms, 105
Common ratio, 265
Commutative law, 2
Completing the square, 206
Complex fractions, 40
Complex numbers, 189
absolute value of, 196
addition and subtraction of, 192
amplitude of, 196
argument of, 196
congugate of, 191
definition of, 189
De Moivre's Theorem, 199
division of, 193, 198
graphical representation of, 195
modulus of, 196
multiplication of, 192, 198
roots of, 200
trigonometric representation
of, 196
Composite number, 25
Composite (reducible) polynomial, 26
Conditional equation, 18
Constant, 53
Constant function, 53
Convergence of series, 259
Coordinate systems
Cartesian, 50
one-dimensional, 49
rectangular, 49
Coordinates, 49, 50
Cosecant, definition of, 65
353
354
Index
Cosine, definition of, 64
Cotangent, definition of, 65
Coterminal angles, 76
Counting numbers, 1
Defective equations, 45
Degree
of a polynomial, 17
relation to radian, 77
of a term, 17
term of highest, 17
unit of angle measure, 77
De Moivre's Theorem, 199
Dependent events, 286
Descartes, Rene, 50
Determinants, 173
expansion of, 176
minors and cof actors of, 175
principal and secondary diagonal
of, 174
properties of, 177
of the second order, 173
solution of linear equations by
means of, 181, 183, 184
sum and product of, 185
of the third order, 175
Directed line segment, 122
Discriminant, 215
Distance between points, 50, 51
Distributive law, 3
Division, 5, 16, 23, 25, 39
Domain of a function, 53
Double-angle formulas, 139
Empirical probability, 284
Equality symbol, 12
Equations
conditional, 18
consistent, 163, 164, 165
defective, 45
definition of, 18
dependent, 165
equivalent, 43
extraneous roots of, 45, 213
inconsistent, 163, 165
independent, 164
linear, 42, 43, 163, 166, 167
in quadratic form, 204, 214
solution of, 42, 163, 171, 181, 183,
184, 212, 224, 227
Exponential equations, 231
Exponents, 16, 86, 88, 90, 92
Factor theorem, 240
Factorial symbol, 97
Factoring, 25, 27
Fractions, 9
addition and substraction of, 36
Fractions (Cont.)
complex, 40
fundamental operations on, 9, 10
multiplication and division of, 39
reduction of, 33
signs associated with, 34
Functions
absolute value, 57
algebraic, 61
classification of, 61
constant, 53
definition of, 53
domain of, 53
exponential, 233
graphs of, 146, 147, 148, 169, 218,
233, 244
greatest integer, 57
identity, 53
inverse, 155
irrational, 62
linear, 54, 169
logarithmic, 233
multiple-valued, 53
notation for, 55
periodic, 149
point, 63
polynomial, 61
range of, 53
rational, 61
rule of correspondence of, 53
single-valued, 53
transcendental, 61
trigonometric, 63
Fundamental assumptions, 1
Fundamental operations, 1, 9, 10, 20
Fundamental theorem of algebra, 241
General term in the binomial
expansion, 99
Graphs
of exponential functions, 233
of inverse trigonometric functions,
158, 159
of linear functions, 169
of logarithmic functions, 233
of polynomials for large values
of x, 244
of trigonometric functions, 146,
147, 148, 149, 151, 152
Greater-than symbol, 12
Greatest common divisor, 30
Greatest-integer function, 57
Half -angle formulas, 140, 300
Highest common factor, 31
Highest degree term, 17
Horizontal line, 50, 169
Identity, 18
Independent events, 286
Index
355
Inequalities, 248
absolute, 248, 254
conditional, 248, 249
involving absolute values, 14
properties of, 248
solution of conditional, 249
Inequality symbol, 12
Initial side, 75
Intercepts, 170
Interpolation, 83, 108
Inverse functions, 155
Irrational functions, 62
Irrational number, 1
Irreducible polynomial, 26
Law of cosines, 296
Law of sines, 289
Law of tangents, 298
Law(s) of exponents, 86
Least common multiple, 32
Less-than symbol, 12
Like terms, 21
Limit of sequence, 258, 259
Line
horizontal, 50, 169
vertical, 51, 169
Linear equation, 42
graphs of, 163, 169, 171
in one unknown, 43
Literal parts, 17
Logarithmic computation, 110
Logarithmic equations, 231
Logarithms, 101
base of, 101
change of base, 113
characteristic of, 106
common, 105
computation by, 110
definition of, 101
laws of, 102
mantissa of, 106, 108
Napierian (natural) ,105
tables of, 108
of trigonometric functions, 131
Mantissa of a logarithm, 106
Mathematical expectation, 284
Mathematical induction, 273
Mean
arithmetic, 262
geometric, 267
harmonic, 264
Meaning of a m/n , 94
Meaning of a, $8
Measurement of angles, 76
Monic polynomial, 18
Most probable number, 284
Multinomial, 17
Multiplication
of algebraic expressions, 22
of fractions, 39
fundamental laws for, 2
Mutually exclusive events, 285
Natural (Napierian) logarithms, 105
Negative exponents, 90
Negative numbers, 7, 8
Number
algebraic, 62
complex, 189
irrational, 1
negative, 1, 7, 8
positive, 1, 7, 8
prime, 25
real, 1
transcendental, 62
Number scale, 6, 49
One-to-one correspondence, 49
Operations
on fractions, 9, 10
with zero, 5
Order of fundamental operations, 20
Order relations for real numbers, 12
Ordered number pairs, 50
Ordinate, 50
Origin, 6, 49
Parentheses, 18
Period, 150
Periodicity, 149, 150
Permutations, 278, 279
Phase, 149, 152
Point function P (t), 63
Polynomial, 17, 18, 26, 61
Positive integer, 1
Positive integral exponents, 16, 8
Positive numbers, 7, 8
Power, 16, 17, 86, 87, 88
Prime, 25, 26
Principal branch, 158
Principal value, 158, 159, 160
Probability, 278
empirical, 284
mathematical, 283
Product, definition of, 2
Product formulas, 143
Progressions, 256
arithmetic, 260
geometric, 265
harmonic, 264
infinite geometric, 268
356
Index
Projections, 122
Quadrant, 49
Quadratic equations
methods for solving, 204, 209
in one unknown, 204
in two unknowns, 221, 224, 227
Quotient, 5
Radian, 77
Radius vector, 52
Range of a function, 53
Rational exponents, 92
Rational function, 61
Rational number, 1
Real number, 1
Reciprocal, 4, 5
Rectangular coordinates, 49, 50
Redundant equation, 45
Remainder theorem, 240
Repeated trials, 287
Repeating decimals, 269
Root of an equation, 42, 45, 213, 217
Scalar quantities, 125
Scientific notation, 92
Secant, definition of, 65
Second degree equation, 204
Segment, line, 122
Sequences, 256, 258, 259
Series, 256, 257, 259
Sexagesimal system, 77
Simultaneous equations, 163, 171,
181, 224, 227
Sine, definition of, 64
Special products, 22
Standard position of an angle, 75
Sum, definition of , 2
Symbols of grouping, 18
Synthetic division, 235
Tables
of logarithms, 108
of trigonometric functions, 71, 2
Tangent, definition of, 64
Terminal side, 75
Terms of algebraic expressions, 2, 17
Theory of equations, 235
Transcendental functions, 62
Triangles, 115
solution of general, 289
solution of right, 117, 133
Trigonometric functions, 63
of angles, 81, 116
definitions of, 64, 65
graphs of, 146, 147, 148, 149,
151, 152
of important special numbers, 65
inverse, 146, 156, 158, 159, 160
logarithms of, 131
of sums and differences, 135
variation of, 146
Trigonometric identities, 68
Variable, 53
dependent, 53
independent, 53
Variation, 57, 58
Vector, 125
components of, 125
magnitude of, 126
multiplication of, by a scalar, 126
normalization of, 127
projection of, 125
representation of, 125, 128
sums and differences of, 127, 129
Vertex of an angle, 75
Vertical lines, 51, 169
#-axis, 49
as-coordinate, 50
y-axis, 49
^/-coordinate, 50
Zero, 3, 5
Zero polynomial, 61
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